# Thread: solving exponents with logs

1. ## solving exponents with logs

Given:
$\displaystyle 5^{x+3} - 5^{x+2} = 2^{3x}$
Find: x

here is my steps

$\displaystyle (x+3)\log{5} - (x+2)\log{5} = (3x)\log{2}$
$\displaystyle \log{5}[(x+3)-(x+2)] = (3x)\log{2}$
$\displaystyle \frac{\log{5}}{\log{2}} = 3x$
$\displaystyle 2.322 = 3x$
$\displaystyle .774 = x$
however this is not the answer.

2. Hello, bigwave!

Given: .$\displaystyle 5^{x+3} - 5^{x+2} \:=\: 2^{3x}$
Find $\displaystyle x.$

Here is my steps

$\displaystyle (x+3)\log{5} - (x+2)\log{5} = (3x)\log{2}$ . . . . wrong!

We must perform some gymnastics first . . .

$\displaystyle \begin{array}{ccccc}\text{We have:} & 5^{x+3} - 5^{x+2} &=& 2^{3x} \\ \\ \text{Factor:} & 5\cdot5^{x+2} - 5^{x+2} &=& 2^{3x} \\ \\ \text{Factor:} & (5-1)5^{x+2} &=& 2^{3x} \\ \\ \text{We have:} & 4\cdot5^{x+2} &=& 2^{3x} \\ \\ \text{Divide by 4:} & 5^{x+2} &=& 2^{3x-2}\end{array}$

Now take logs: .$\displaystyle \ln\left(5^{x+2}\right) \;=\;\ln\left(2^{3x-2}\right) \quad\Rightarrow\quad (x+2)\!\cdot\!\ln(5) \;=\;(3x-2)\!\cdot\!\ln(2)$

. . and so on . . .

3. Originally Posted by Soroban
Now take logs: .$\displaystyle \ln\left(5^{x+2}\right) \;=\;\ln\left(2^{3x-2}\right) \quad\Rightarrow\quad (x+2)\!\cdot\!\ln(5) \;=\;(3x-2)\!\cdot\!\ln(2)$
as suggested...

$\displaystyle (x+2)(1.609)=(3x-2)(.693)$
$\displaystyle 1.609x+3.218=2.079x-1.386$
$\displaystyle -.470x = -4.604$
$\displaystyle x = 9.8$