{"id":1164,"date":"2025-07-23T17:05:15","date_gmt":"2025-07-23T17:05:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1164"},"modified":"2026-03-18T19:33:10","modified_gmt":"2026-03-18T19:33:10","slug":"exponential-and-logarithmic-models-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models-learn-it-3\/","title":{"raw":"Exponential and Logarithmic Equations: Learn It 3","rendered":"Exponential and Logarithmic Equations: Learn It 3"},"content":{"raw":"

Logarithmic Equations<\/h2>\r\nWe have already seen that every logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equal to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.\r\n\r\n
For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:\r\n

[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Convert to exponential form}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/section>

\r\n

using the definition of a logarithm to solve logarithmic equations<\/h3>\r\nFor any algebraic expression [latex]S[\/latex] and real numbers [latex]b[\/latex] and [latex]c[\/latex], where [latex]b>0,\\text{ }b\\ne 1[\/latex],\r\n

[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\r\n\r\n<\/section>

a) Solve [latex]2\\mathrm{ln}x+3=7[\/latex].[reveal-answer q=\"977891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977891\"]\r\n

[latex]\\begin{array}{l}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{}2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3 from both sides}.\\hfill \\\\ \\text{}\\mathrm{ln}x=2\\hfill & \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}x={e}^{2}\\hfill & \\text{Rewrite in exponential form}.\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\nb) Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].[reveal-answer q=\"231886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"231886\"]\r\n

[latex]\\begin{array}{l}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{}\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide both sides by 2}.\\hfill \\\\ \\text{}6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{}x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

As was the case when using the properties and rules of exponents and logarithms to rewrite expressions containing them, there can be more than one good way to solve a logarithmic equation. It is good practice to follow the examples given for each of the situations in this section, but you should think about alternative ways to creatively and correctly apply the properties and rules.<\/section>
[ohm_question hide_question_numbers=1]321539[\/ohm_question]<\/section>
[ohm_question hide_question_numbers=1]321540[\/ohm_question]<\/section><\/section>\r\n
\r\n
[ohm_question hide_question_numbers=1]321541[\/ohm_question]<\/section><\/div>\r\n<\/div>","rendered":"

Logarithmic Equations<\/h2>\n

We have already seen that every logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equal to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\n

For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:<\/p>\n

[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Convert to exponential form}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n

\n

using the definition of a logarithm to solve logarithmic equations<\/h3>\n

For any algebraic expression [latex]S[\/latex] and real numbers [latex]b[\/latex] and [latex]c[\/latex], where [latex]b>0,\\text{ }b\\ne 1[\/latex],<\/p>\n

[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\n<\/section>\n

a) Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\n