# Thread: Two Exponential Problems

1. ## Two Exponential Problems

I don't know how to solve these two problems:

1.) $100 = 5000(0.02^{0.5^x})$

2.) $3^x(1/9)^{x+1} = 27(3^{-x})$

Your assistance is much appreciated.

2. #1: I am going to use fractions instead of decimals.

$\frac{1}{50}=(\frac{1}{50})^{\frac{1}{2}x}$

log of both sides:

$log(\frac{1}{50})=\frac{1}{2}xlog(\frac{1}{50})$

$1=\frac{1}{2}x$

$x=2$

Use your log properties.

3. For #2

$
3^x(1/9)^{x+1} = 27(3^{-x})
$

Since 9 and 27 are both powers of 3 rewrite them as their prime factorization.

$\frac{1}{9}=3^{-2}$ and $27=3^3$

Then we get

$3^x \cdot (3^{-2})^{x+1}=3^3 \cdot(3^{-x})$

Using exponent rules we get

$3^x \cdot (3^{-2x-2})=(3^{-x+3})$

$(3^{-x-2})=(3^{-x+3})$

since the bases are the same the exponents must be equal so

$-x+2=-x+3$ this equation has no solutions. Therefore this equation has no solutions.