# Solving Exponential Equations

This post will outline different types of solving exponential equations and how to treat each one of them. Before reading this topic it is extremely important that we already know the law of exponent.  So that we will be able to know when to add, subtract, multiply and divide the exponent. After this topic we will discuss one of the most useful applications of exponential equations, the half-life and growth and how to solve them.

Sample Problem 1:

Solve for x: $4^{x-2}=2^{x+5}$

[toggle title=”Solution”]

Express both side to base 2.

$(2^2)^{x-2}=2^{x+5}$ $2^{2x-4}=2^{x+5}$ $2x-4=x+5$

$x=9$ [/toggle]

Sample Problem 2:

Solve for x:

$\displaystyle\frac{3\cdot 9^{2-x}}{3^x}=3^{x+1}$

[toggle title=”Solution”]

$(\displaystyle\frac{3\cdot 9^{2-x}}{3^x}=3^{x+1})3^x$ $3\cdot 9^{2-x}=3^{x+1}(3^x)$ $3\cdot (3^2)^{2-x}=3^{x+1}(3^x)$ $3(3^{4-2x})=3^{2x+1}$ $3^{5-2x}=3^{2x+1}$ $5-2x=2x+1$ $4x=4$

$x=1$ [/toggle]

Sample Problem 3:

Solve for x: $(4^x)^x=\displaystyle\frac{64^x}{16}$

[toggle title=”Solution”]

Express all bases in 4 or 2.

$4^{x^2}=\displaystyle\frac{4^{3x}}{4^2}$ $4^{x^2}=4^{3x-2}$ $x^2=3x-2$ $x^2-3x+2=0$ $(x-1)(x-2)=0$

$x=1$ or $x=2$

By quick check both values of x satisfy the original equation. [/toggle]

Sample Problem 4:

Solve for x: $3^x=6^x$

[toggle title=”Solution 1″]

This is a trivial equation, by quick inspection we know that any number raised to 0 is 1. Making zero is the root of equation. [/toggle]

[toggle title=”Solution 2″]

By logarithm

$3^x=6^x$ $x\log 3=x\log 6$ $x\log 3-x\log 6=0$ $x(\log 3-\log 6)=0$

$x=0$ [/toggle]

Sample Problem 5:

Solve for x: $3^x=4^{x-1}$

[toggle title=”Solution”]

The bases has cannot be simplified to two common bases

By applying log both sides

$\log 3^x=\log 4^{x-1}$ $x\log 3=(x-1)\log 4$ $x\log 3=x\log 4-\log 4$ $x\log 3-x\log 4=-\log 4$ $x(\log 3-\log 4)=-\log 4$

$x=\displaystyle\frac{-\log 4}{\log 3-\log 4}$ [/toggle]

Sample Problem 6:

Solve for x: $3^{2x-5}=5^{4-x}$

[toggle title=”Solution”]

By applying log both sides

$\log 3^{2x-5}=\log 5^{4-x}$ $(2x-5)\log 3=(4-x ) \log 5$ $2x\log 3-5\log 3=4\log 5-x\log 5$ $2x\log 3+x\log 5=4\log 5+5\log 3$ $x(2\log 3+\log 5)=4\log 5+5\log 3$

$x=\displaystyle\frac{4\log 5+5\log 3}{2\log 3+\log 5}$ [/toggle]

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.