Examples of Solving Exponential Equations

Example

Solve $7^x=12.\:$Give the solution to the nearest thousandth.

Solution

The properties of exponents cannot be used to solve this equation, so we apply the useful property of logarithms on the previous page. While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here.

$7^x=12$

$\ln7^x=\ln12$

$x\:\ln7=\ln12\\ x = \frac{\ln12}{\ln7}\\ x\approx 1.277$

The solution set is {1.277}.

NOTE: When evaluating a quotient of logarithms like $\frac{\ln12}{\ln7},\:$do not confuse this quotient with $\ln\frac{12}{7}\:$(which is equivalent to $\ln12-\ln7$).

We cannot change the quotient of two logarithms to a difference of logarithms.

$\frac{\ln12}{\ln7}\ne\ln\frac{12}{7}$

Example

Solve $3^{2x-1}=0.4^{x+2}$. Give the solution to the nearest thousandth.

Solution

$3^{2x-1}=0.4^{x+2}\\ ln3^{2x-1} =ln0.4^{x+2} \\ (2x-1)ln3=(x+2)ln0.4\\ 2xln3-ln3=xln0.4+2ln0.4\\ 2xln3-xln0.4=2ln0.4+ln3\\ x(2ln3-ln0.4)=2ln0.4+ln3\\ x=\frac{ln0.4^{2} +ln3}{ln3^{2}-ln0.4 } \\ x=\frac{ln0.16+ln3}{ln9-ln0.4} \\ x=\frac{ln0.48}{ln22.5}\\ x\approx -0.236$

The solution set is {-0.236}.

Examples To Try

Below are several more examples. Write them down and then try to solve them on your own before reading the solutions provided. (If you need to, return to the page showing all of the logarithm properties while you work on the problems.)

1. $e^{x^2}=200$
2. $e^{2x+1}\cdot e^{-4x}=3e$
3. $e^{2x}-4e^x+3=0$

Solutions

1. $e^{x^2}=200$
   Take the natural logarithm of both sides.
   $\ln e^{x^2}=\ln200$
   Use the power property for logarithms and the fact that $\ln e=1$.
   $x^2=\ln200$
   Take the square root of both sides, and introduce a $\pm$ to get both roots.
   $x=\pm\sqrt[]{\ln200}\approx\pm2.302$
   The solution set is {$\pm2.302$}.
2. $e^{2x+1}\cdot e^{-4x}=3e$
   Use the exponent property $a^na^m=a^{n+m}$.
   $e^{-2x+1}=3e$
   Divide by $e.$
   $e^{-2x}=3$
   Take the natural logarithm of both sides and use the power property.
   $\ln e^{-2x}=\ln3\\ -2xlne=ln3\\ -2x=ln3\\ x=-\frac{1}{2} ln3\approx -0.549$
   The solution set is {-0.549}
3. $e^{2x}-4e^x+3=0$
   If we substitute $u=e^x,\:$we notice that the equation is quadratic in form.
   $e^{2x}-4e^x+3=0$
   $\left(e^x\right)^2-4\left(e^x\right)+3=0\\u^{2} -4u+3=0\\ (u-1)(u-3)=0$
   $u=1$ or $u=3$
   Now substitute $e^{x}$ for $u,$ and solve.
   $e^x=1\:or\:e^x=3\\ lne^{x} =ln1\:or\:lne^{x}=ln3\\ x=0\:or\:x=ln3$
   Both values check, so the solution set is {0, ln 3}.