# 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.

### 5 Responses

1. Jame says:

hi!,I love your writing very much! percentage we communicate more about your article on AOL? I require a specialist on this house to unravel my problem. Maybe that is you! Taking a look forward to see you.

2. Normally I do not read post on blogs, but I would like to say that this write-up very forced me to try and do so! Your writing style has been surprised me. Thanks, quite nice article.

3. Kena Soukup says:

I’m also writing to let you know of the cool experience my daughter had checking yuor web blog. She learned several issues, including what it’s like to possess an amazing giving mindset to get most people without difficulty fully understand chosen tortuous issues. You really exceeded people’s expected results. Thank you for churning out such great, safe, educational and also unique tips on the topic.

4. Excellent publish from specialist also it will probably be a fantastic know how to me and thanks really much for posting this helpful data with us all.

5. I’m also writing to let you know of the cool experience my daughter had checking yuor web blog. She learned several issues, including what it’s like to possess an amazing giving mindset to get most people without difficulty fully understand chosen tortuous issues. You really exceeded people’s expected results. Thank you for churning out such great, safe, educational and also unique tips on the topic.