{"id":1274,"date":"2024-04-15T18:15:45","date_gmt":"2024-04-15T18:15:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1274"},"modified":"2024-08-04T12:17:59","modified_gmt":"2024-08-04T12:17:59","slug":"exponential-and-logarithmic-functions-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/exponential-and-logarithmic-functions-learn-it-5\/","title":{"raw":"Exponential and Logarithmic Functions: Learn It 5","rendered":"Exponential and Logarithmic Functions: Learn It 5"},"content":{"raw":"<h2>Hyperbolic Functions Cont.<\/h2>\r\n<h3>Identities Involving Hyperbolic Functions<\/h3>\r\n<p>Just as trigonometric functions have identities that allow for the simplification and transformation of expressions, hyperbolic functions also possess their own set of identities.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>hyperbolic function identities<\/h3>\r\n<div><strong>Hyperbolic Reflection Identities:<\/strong>\r\n<ul>\r\n\t<li>[latex]\\cosh(\u2212x)=\\cosh x[\/latex]<\/li>\r\n\t<li>[latex]\\sinh(\u2212x)=\u2212\\sinh x[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Hyperbolic Pythagorean Identities:<\/strong>\r\n<ul>\r\n\t<li>[latex]\\cosh^2 x-\\sinh^2 x=1[\/latex]<\/li>\r\n<\/ul>\r\n<p><strong>Hyperbolic Squared Identities:<\/strong><\/p>\r\n<ul>\r\n\t<li>[latex]1-\\tanh^2 x=\\text{sech}^2 x[\/latex]<\/li>\r\n\t<li>[latex]\\coth^2 x-1=\\text{csch}^2 x[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Hyperbolic Addition Formulas:<\/strong>\r\n<ul>\r\n\t<li>[latex]\\sinh(x \\pm y)=\\sinh x \\cosh y \\pm \\cosh x \\sinh y[\/latex]<\/li>\r\n\t<li>[latex]\\cosh (x \\pm y)=\\cosh x \\cosh y \\pm \\sinh x \\sinh y[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Exponential Definitions<\/strong> <strong>of Hyperbolic Functions<\/strong><br \/>\r\n<ul>\r\n\t<li>[latex]\\cosh x+\\sinh x=e^x[\/latex]<\/li>\r\n\t<li>[latex]\\cosh x-\\sinh x=e^{\u2212x}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<ol id=\"fs-id1170572443403\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Simplify [latex]\\sinh(5 \\ln x)[\/latex].<\/li>\r\n\t<li>If [latex]\\sinh x=\\frac{3}{4}[\/latex], find the values of the remaining five hyperbolic functions.<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"fs-id1170572443462\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572443462\"]<\/p>\r\n<ol id=\"fs-id1170572443462\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>Using the definition of the [latex]\\sinh[\/latex] function, we write<br \/>\r\n<div id=\"fs-id1170570995857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sinh(5 \\ln x)=\\large \\frac{e^{5 \\ln x}-e^{-5 \\ln x}}{2} \\normalsize = \\large \\frac{e^{\\ln(x^5)}-e^{\\ln(x^{-5})}}{2} \\normalsize =\\large \\frac{x^5-x^{-5}}{2}[\/latex].<\/div>\r\n<\/li>\r\n\t<li>Using the identity [latex]\\cosh^2 x-\\sinh^2 x=1[\/latex], we see that<br \/>\r\n<div id=\"fs-id1170573388429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cosh^2 x=1+\\big(\\frac{3}{4}\\big)^2=\\frac{25}{16}[\/latex].<\/div>\r\n<p>Since [latex]\\cosh x \\ge 1[\/latex] for all [latex]x[\/latex], we must have [latex]\\cosh x=5\/4[\/latex]. Then, using the definitions for the other hyperbolic functions, we conclude that [latex]\\tanh x=3\/5, \\, \\text{csch} \\, x=4\/3, \\, \\text{sech} \\, x=4\/5[\/latex], and [latex]\\coth x=5\/3[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n\r\n\r\nWatch the following video to see the worked solution to this example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=1498&amp;end=1738&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\n\r\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.5ExponentialAndLogarithmicFunctions1498to1738_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h3>Inverse Hyperbolic Functions<\/h3>\r\n<p id=\"fs-id1170572549987\">From the graphs of the hyperbolic functions, we see that all of them are one-to-one except [latex]\\cosh x[\/latex] and [latex]\\text{sech} \\, x[\/latex]. If we restrict the domains of these two functions to the interval [latex][0,\\infty)[\/latex], then all the hyperbolic functions are one-to-one, and we can define the <strong>inverse hyperbolic functions<\/strong>. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>inverse hyperbolic functions<\/h3>\r\n<div>[latex]\\begin{array}{cccc}\\sinh^{-1} x=\\text{arcsinh } x=\\ln(x+\\sqrt{x^2+1})\\hfill &amp; &amp; &amp; \\cosh^{-1} x=\\text{arccosh } x=\\ln(x+\\sqrt{x^2-1})\\hfill \\\\ \\tanh^{-1} x=\\text{arctanh } x=\\frac{1}{2}\\ln\\big(\\frac{1+x}{1-x}\\big)\\hfill &amp; &amp; &amp; \\coth^{-1} x=\\text{arccot } x=\\frac{1}{2}\\ln\\big(\\frac{x+1}{x-1}\\big)\\hfill \\\\ \\text{sech}^{-1} x=\\text{arcsech } x=\\ln\\big(\\frac{1+\\sqrt{1-x^2}}{x}\\big)\\hfill &amp; &amp; &amp; \\text{csch}^{-1} x=\\text{arccsch } x=\\ln\\big(\\frac{1}{x}+\\frac{\\sqrt{1+x^2}}{|x|}\\big)\\hfill \\end{array}[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox questionHelp\">\r\n<p id=\"fs-id1170572235136\">Let\u2019s look at how to derive the first equation, [latex]\\sinh^{-1} x=\\text{arcsinh } x=\\ln(x+\\sqrt{x^2+1})[\/latex]. The others follow similarly.<\/p>\r\n<p>Suppose [latex]y=\\sinh^{-1} x[\/latex]. Then, [latex]x=\\sinh y[\/latex] and, by the definition of the hyperbolic sine function, [latex]x=\\frac{e^y-e^{\u2212y}}{2}[\/latex].\u00a0<br \/>\r\n<br \/>\r\nMultiplying both sides of this equation by [latex]2[\/latex] to get rid of the fraction and setting that equal to zero, we get<\/p>\r\n<div id=\"fs-id1170572235202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^y-2x-e^{\u2212y}=0[\/latex]<\/div>\r\n<p>Multiplying this equation by [latex]e^y[\/latex], we obtain<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2y}-2xe^y-1=0[\/latex]<\/div>\r\n<p id=\"fs-id1170572235286\">This can be solved like a quadratic equation, with the solution<\/p>\r\n<div id=\"fs-id1170572235290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^y=\\large \\frac{2x \\pm \\sqrt{4x^2+4}}{2} \\normalsize =x \\pm \\sqrt{x^2+1}[\/latex]<\/div>\r\n<p id=\"fs-id1170572482176\">Since [latex]e^y&gt;0[\/latex], the only solution is the one with the positive sign.<\/p>\r\n<p>Applying the natural logarithm to both sides of the equation, we conclude that<\/p>\r\n<div id=\"fs-id1170572482196\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\ln(x+\\sqrt{x^2+1})[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Evaluate each of the following expressions<\/p>\r\n<p>[latex]\\sinh^{-1}(2)[\/latex]<br \/>\r\n[latex]\\tanh^{-1}\\left(\\frac{1}{4}\\right)[\/latex]<\/p>\r\n<p>[reveal-answer q=\"277655\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"277655\"]<\/p>\r\n<p id=\"fs-id1170572482306\">[latex]\\sinh^{-1}(2)=\\ln(2+\\sqrt{2^2+1})=\\ln(2+\\sqrt{5}) \\approx 1.4436[\/latex]<\/p>\r\n<p id=\"fs-id1170572176029\">[latex]\\tanh^{-1}(\\frac{1}{4})=\\frac{1}{2}\\ln(\\frac{1+1\/4}{1-1\/4})=\\frac{1}{2}\\ln(\\frac{5\/4}{3\/4})=\\frac{1}{2}\\ln(\\frac{5}{3}) \\approx 0.2554[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>Hyperbolic Functions Cont.<\/h2>\n<h3>Identities Involving Hyperbolic Functions<\/h3>\n<p>Just as trigonometric functions have identities that allow for the simplification and transformation of expressions, hyperbolic functions also possess their own set of identities.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>hyperbolic function identities<\/h3>\n<div><strong>Hyperbolic Reflection Identities:<\/strong><\/p>\n<ul>\n<li>[latex]\\cosh(\u2212x)=\\cosh x[\/latex]<\/li>\n<li>[latex]\\sinh(\u2212x)=\u2212\\sinh x[\/latex]<\/li>\n<\/ul>\n<p><strong>Hyperbolic Pythagorean Identities:<\/strong><\/p>\n<ul>\n<li>[latex]\\cosh^2 x-\\sinh^2 x=1[\/latex]<\/li>\n<\/ul>\n<p><strong>Hyperbolic Squared Identities:<\/strong><\/p>\n<ul>\n<li>[latex]1-\\tanh^2 x=\\text{sech}^2 x[\/latex]<\/li>\n<li>[latex]\\coth^2 x-1=\\text{csch}^2 x[\/latex]<\/li>\n<\/ul>\n<p><strong>Hyperbolic Addition Formulas:<\/strong><\/p>\n<ul>\n<li>[latex]\\sinh(x \\pm y)=\\sinh x \\cosh y \\pm \\cosh x \\sinh y[\/latex]<\/li>\n<li>[latex]\\cosh (x \\pm y)=\\cosh x \\cosh y \\pm \\sinh x \\sinh y[\/latex]<\/li>\n<\/ul>\n<p><strong>Exponential Definitions<\/strong> <strong>of Hyperbolic Functions<\/strong><\/p>\n<ul>\n<li>[latex]\\cosh x+\\sinh x=e^x[\/latex]<\/li>\n<li>[latex]\\cosh x-\\sinh x=e^{\u2212x}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<ol id=\"fs-id1170572443403\" style=\"list-style-type: lower-alpha;\">\n<li>Simplify [latex]\\sinh(5 \\ln x)[\/latex].<\/li>\n<li>If [latex]\\sinh x=\\frac{3}{4}[\/latex], find the values of the remaining five hyperbolic functions.<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572443462\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572443462\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572443462\" style=\"list-style-type: lower-alpha;\">\n<li>Using the definition of the [latex]\\sinh[\/latex] function, we write\n<div id=\"fs-id1170570995857\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sinh(5 \\ln x)=\\large \\frac{e^{5 \\ln x}-e^{-5 \\ln x}}{2} \\normalsize = \\large \\frac{e^{\\ln(x^5)}-e^{\\ln(x^{-5})}}{2} \\normalsize =\\large \\frac{x^5-x^{-5}}{2}[\/latex].<\/div>\n<\/li>\n<li>Using the identity [latex]\\cosh^2 x-\\sinh^2 x=1[\/latex], we see that\n<div id=\"fs-id1170573388429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cosh^2 x=1+\\big(\\frac{3}{4}\\big)^2=\\frac{25}{16}[\/latex].<\/div>\n<p>Since [latex]\\cosh x \\ge 1[\/latex] for all [latex]x[\/latex], we must have [latex]\\cosh x=5\/4[\/latex]. Then, using the definitions for the other hyperbolic functions, we conclude that [latex]\\tanh x=3\/5, \\, \\text{csch} \\, x=4\/3, \\, \\text{sech} \\, x=4\/5[\/latex], and [latex]\\coth x=5\/3[\/latex].<\/p>\n<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tOkk_pSFpzk?controls=0&amp;start=1498&amp;end=1738&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the transcript for this video using <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.5ExponentialAndLogarithmicFunctions1498to1738_transcript.txt\" target=\"_blank\" rel=\"noopener\">this link<\/a> (opens in new window).<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h3>Inverse Hyperbolic Functions<\/h3>\n<p id=\"fs-id1170572549987\">From the graphs of the hyperbolic functions, we see that all of them are one-to-one except [latex]\\cosh x[\/latex] and [latex]\\text{sech} \\, x[\/latex]. If we restrict the domains of these two functions to the interval [latex][0,\\infty)[\/latex], then all the hyperbolic functions are one-to-one, and we can define the <strong>inverse hyperbolic functions<\/strong>. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>inverse hyperbolic functions<\/h3>\n<div>[latex]\\begin{array}{cccc}\\sinh^{-1} x=\\text{arcsinh } x=\\ln(x+\\sqrt{x^2+1})\\hfill & & & \\cosh^{-1} x=\\text{arccosh } x=\\ln(x+\\sqrt{x^2-1})\\hfill \\\\ \\tanh^{-1} x=\\text{arctanh } x=\\frac{1}{2}\\ln\\big(\\frac{1+x}{1-x}\\big)\\hfill & & & \\coth^{-1} x=\\text{arccot } x=\\frac{1}{2}\\ln\\big(\\frac{x+1}{x-1}\\big)\\hfill \\\\ \\text{sech}^{-1} x=\\text{arcsech } x=\\ln\\big(\\frac{1+\\sqrt{1-x^2}}{x}\\big)\\hfill & & & \\text{csch}^{-1} x=\\text{arccsch } x=\\ln\\big(\\frac{1}{x}+\\frac{\\sqrt{1+x^2}}{|x|}\\big)\\hfill \\end{array}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p id=\"fs-id1170572235136\">Let\u2019s look at how to derive the first equation, [latex]\\sinh^{-1} x=\\text{arcsinh } x=\\ln(x+\\sqrt{x^2+1})[\/latex]. The others follow similarly.<\/p>\n<p>Suppose [latex]y=\\sinh^{-1} x[\/latex]. Then, [latex]x=\\sinh y[\/latex] and, by the definition of the hyperbolic sine function, [latex]x=\\frac{e^y-e^{\u2212y}}{2}[\/latex].\u00a0<\/p>\n<p>Multiplying both sides of this equation by [latex]2[\/latex] to get rid of the fraction and setting that equal to zero, we get<\/p>\n<div id=\"fs-id1170572235202\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^y-2x-e^{\u2212y}=0[\/latex]<\/div>\n<p>Multiplying this equation by [latex]e^y[\/latex], we obtain<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^{2y}-2xe^y-1=0[\/latex]<\/div>\n<p id=\"fs-id1170572235286\">This can be solved like a quadratic equation, with the solution<\/p>\n<div id=\"fs-id1170572235290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e^y=\\large \\frac{2x \\pm \\sqrt{4x^2+4}}{2} \\normalsize =x \\pm \\sqrt{x^2+1}[\/latex]<\/div>\n<p id=\"fs-id1170572482176\">Since [latex]e^y>0[\/latex], the only solution is the one with the positive sign.<\/p>\n<p>Applying the natural logarithm to both sides of the equation, we conclude that<\/p>\n<div id=\"fs-id1170572482196\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\ln(x+\\sqrt{x^2+1})[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Evaluate each of the following expressions<\/p>\n<p>[latex]\\sinh^{-1}(2)[\/latex]<br \/>\n[latex]\\tanh^{-1}\\left(\\frac{1}{4}\\right)[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q277655\">Show Solution<\/button><\/p>\n<div id=\"q277655\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572482306\">[latex]\\sinh^{-1}(2)=\\ln(2+\\sqrt{2^2+1})=\\ln(2+\\sqrt{5}) \\approx 1.4436[\/latex]<\/p>\n<p id=\"fs-id1170572176029\">[latex]\\tanh^{-1}(\\frac{1}{4})=\\frac{1}{2}\\ln(\\frac{1+1\/4}{1-1\/4})=\\frac{1}{2}\\ln(\\frac{5\/4}{3\/4})=\\frac{1}{2}\\ln(\\frac{5}{3}) \\approx 0.2554[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":21,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":759,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1274"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":14,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1274\/revisions"}],"predecessor-version":[{"id":1295,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1274\/revisions\/1295"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/759"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1274\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1274"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1274"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1274"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1274"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}