Chapter 3.4, Problem 37E

To determine

To find : solution to the exponential equation algebraically.

Answer to Problem 37E

The solution to the equation $3^x=2^{x-1}$ is $\underset{\_}{\boxed{x\approx -1.710}}$

Explanation of Solution

Given information: $3^x=2^{x-1}$

Concept Involved:

The word “solve” means the process of finding the value of x that makes the equation true. In order to solve we need to undo whatever is done to x .

Formula Used

  $\ln m^n=n\left(\ln m\right)$

Calculation:

Take natural logarithm on both sides of the equation

  $\ln 3^x=\ln 2^{x-1}$

Use the logarithmic property $\ln m^n=n\left(\ln m\right)$ on both sides of the equation to rewrite it

  $x\left(\ln 3\right)=\left(x-1\right)\cdot \ln 2$

Distribute ln2 in the right side of the equation

  $x\cdot \ln 3=x\cdot \ln 2-\ln 2$

Subtracting xln2 on both sides of the equation in order to get x alone in one side of the equation

  $x\cdot \ln 3-x\cdot \ln 2=x\cdot \ln 2-\ln 2-x\cdot \ln 2$

Simplify on both sides of the equation

  $x\cdot \ln 3-x\cdot \ln 2=-\ln 2$

Factor x in left side of the equation

  $x\cdot \left(\ln 3-\ln 2\right)=-\ln 2$

Divide $\ln 3-\ln 2$ on both sides of the equation

  $\frac{x\cdot \left(\ln 3-\ln 2\right)}{\left(\ln 3-\ln 2\right)}=\frac{-\ln 2}{\left(\ln 3-\ln 2\right)}$

Simplify fraction on both sides of the equation

  $x\approx -1.70951129$

Check this in the original equation.

  $x\approx -1.70951129$ makes the original equation TRUE

Conclusion:

The solution to the equation $3^x=2^{x-1}$ is $\underset{\_}{\boxed{x\approx -1.710}}$