Introduction to Differential Equations: Background You’ll Need 4
Use logarithms to solve exponential equations
Logarithmic Functions
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.
The exponential function [latex]f(x)=b^x[/latex] is one-to-one, with domain [latex](−\infty ,\infty)[/latex] and range [latex](0,\infty )[/latex]. Therefore, it has an inverse function, called the logarithmic function with base [latex]b[/latex].
For any [latex]b>0, \, b \ne 1[/latex], the logarithmic function with base [latex]b[/latex], denoted [latex]\log_b[/latex], has domain [latex](0,\infty )[/latex] and range [latex](−\infty ,\infty )[/latex], and satisfies
[latex]\log_b(x)=y[/latex] if and only if [latex]b^y=x[/latex].
logarithmic functions
A logarithmic function is the inverse of an exponential function and is written as [latex]log_{b}(x)[/latex]. For a given base [latex]b[/latex], it tells us the power to which [latex]b[/latex] must be raised to get [latex]x[/latex].
The most commonly used logarithmic function is the function [latex]\log_e (x)[/latex]. Since this function uses natural [latex]e[/latex] as its base, it is called the natural logarithm. Here we use the notation [latex]\ln(x)[/latex] or [latex]\ln x[/latex] to mean [latex]\log_e (x)[/latex].
Solve each of the following equations for [latex]x[/latex].
[latex]\ln \left(\frac{1}{x}\right)=4[/latex]
[latex]\log_{10} \sqrt{x}+ \log_{10} x=2[/latex]
[latex]\ln(2x)-3 \ln(x^2)=0[/latex]
By the definition of the natural logarithm function,
[latex]\ln\big(\frac{1}{x}\big)=4 \, \text{ if and only if } \, e^4=\frac{1}{x}[/latex]
Therefore, the solution is [latex]x=\frac{1}{e^4}[/latex].
Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
[latex]\log_{10} \sqrt{x}+ \log_{10} x = \log_{10} x \sqrt{x} = \log_{10}x^{3/2} = \frac{3}{2} \log_{10} x[/latex]
Therefore, the equation can be rewritten as
[latex]\frac{3}{2} \log_{10} x = 2 \, \text{ or } \, \log_{10} x = \frac{4}{3}[/latex]
The solution is [latex]x=10^{4/3}=10\sqrt[3]{10}[/latex].
Using the power property of logarithmic functions, we can rewrite the equation as [latex]\ln(2x) - \ln(x^6) = 0[/latex].
Using the quotient property, this becomes
[latex]\ln\big(\frac{2}{x^5}\big)=0[/latex]
Therefore, [latex]\frac{2}{x^5}=1[/latex], which implies [latex]x=\sqrt[5]{2}[/latex]. We should then check for any extraneous solutions.
Solve each of the following equations for [latex]x[/latex].
[latex]5^x=2[/latex]
[latex]e^x+6e^{−x}=5[/latex]
Applying the natural logarithm function to both sides of the equation, we have
[latex]\ln 5^x=\ln 2[/latex]
Using the power property of logarithms,
[latex]x \ln 5=\ln 2[/latex]
Therefore, [latex]x=\frac{\ln 2 }{\ln 5}[/latex].
Multiplying both sides of the equation by [latex]e^x[/latex], we arrive at the equation
[latex]e^{2x}+6=5e^x[/latex]
Rewriting this equation as
[latex]e^{2x}-5e^x+6=0[/latex],
we can then rewrite it as a quadratic equation in [latex]e^x[/latex]:
[latex](e^x)^2-5(e^x)+6=0[/latex]
Now we can solve the quadratic equation. Factoring this equation, we obtain
[latex](e^x-3)(e^x-2)=0[/latex]
Therefore, the solutions satisfy [latex]e^x=3[/latex] and [latex]e^x=2[/latex]. Taking the natural logarithm of both sides gives us the solutions
[latex]x=\ln 3, \, \ln 2[/latex]
Watch the following video to see the worked solution to this example.
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