# Thread: Help with solving exponential equation

1. ## Help with solving exponential equation

Solve the following equation for 'm':

(1/5)^m * (1/4)^18 = 1/(2*(10)^35)

I don't remember how to solve this, and I need help. A step by step solution would be very helpful.

Thanks!

2. $\displaystyle \left(\frac{1}{5}\right)^m = \frac{1^m}{5^m} = \frac{1}{5^m}$

so

$\displaystyle \frac{1}{5^m}\times \left(\frac{1}{4}\right)^{18} =\frac{1}{2\times (10)^{35}}$

$\displaystyle \frac{1}{5^m}\times \frac{1}{4^{18}} =\frac{1}{2\times (10)^{35}}$

$\displaystyle 5^m = \frac{2\times (10)^{35}}{4^{18}}\implies m= \dots$

3. So here's the thing: I have been out of school for several years, and I'm thinking about going back to get a Master's Degree. I was taking a practice GMAT test and this was one of the questions. I can do this problem with a calculator, but I am supposed to be able to do this by hand for the test. I understand everything you've done so far, but I still don't know how to do this by hand. I am pretty sure once I see this worked out that it will come back to me.

4. I have done it all bar the last step.

It is taking $log_5$ of both sides.

5. $10= 2(5)$ and $4= 2(2)$ so you can write everything on the right as powers of 2 and 5.

Do it! You'll like it!

6. $\displaystyle 5^m = \frac{2 \cdot 10^{35}}{4^{18}}$

$\displaystyle 5^m = \frac{2 \cdot 10^{35}}{2^{36}}$

$\displaystyle 5^m = \frac{10^{35}}{2^{35}} = 5^{35}$

7. Thanks so much for the help. You guys are the best!

8. 1/5)^m = 1/5^m, and (1/4)^18 = 1/4^18 (since 1 raised to any power = 1)
Additionally, 4^18 = (2^2)^18 = 2^36.

Since both sides of the equation have 1 in the their numerators, to set them = we have to set the demoninators equal to each other.

The denominator of the left side is now = 5^m*2^36

The denominator of the right side = 2(10)^35, which can be rewritten as (2)(2*5)^35, or (2)(2^35)(5^35).
Since (2)(2^35)= 2^36, 2^36 can be cancelled from each side of the equation. We are now left with 5^m = 5^35. m = 35, and the correct answer is D