Chapter 6.6, Problem 73E

To determine

To describe: The general method for solving exponential and logarithmic equations with different bases.

Answer to Problem 73E

Both types of equations can be solved graphically with ease, but algebraically the solution will be a little bit tedious.

Explanation of Solution

Given information: Exponential and logarithmic equations are with different bases.

Calculation: Suppose the given equations are in the general form that is $a^{f\left(x\right)}=b^{g\left(x\right)},a\ne b$

Take logarithm both sides to solve,

  $\begin{array}{l}\log a^{f\left(x\right)}=\log b^{g\left(x\right)}\\ f\left(x\right)\log a=g\left(x\right)\log b\\ \frac{f\left(x\right)}{g\left(x\right)}=\frac{\log b}{\log a}\end{array}$

Logarithmic equations having different bases, ${\log }_{a}f\left(x\right)={\log }_{b}g\left(x\right),a\ne b$

By using change-of-base property,

  ${\log }_{a}f\left(x\right)=\frac{{\log }_{a}g\left(x\right)}{{\log }_{a}b}$

This equation is tedious to solve unless $f\left(x\right)$ and $g\left(x\right)$ have friendly expression, this type of case we have in question number $70$ .

  $\begin{array}{l}{\log }_{4}x={\log }_{8}4x\\ {\log }_{4}x=\frac{{\log }_{4}4x}{{\log }_{4}8}\\ {\log }_{4}x=\frac{{\log }_{4}4+{\log }_{4}x}{{\log }_{4}8}\\ {\log }_{4}x=\frac{1+{\log }_{4}x}{{\log }_{4}8}\end{array}$

  $\begin{array}{l}{\log }_{4}x{\log }_{4}8=1+{\log }_{4}x\\ {\log }_{4}x{\log }_{4}8-{\log }_{4}x=1\\ {\log }_{4}x\left({\log }_{4}8-1\right)=1\\ {\log }_{4}x=\frac{1}{{\log }_{4}8-1}\end{array}$

Mostly, the graphical intersection of equations gives the solution, and it is easy too.