I don't know how to solve these two problems:
1.) $\displaystyle 100 = 5000(0.02^{0.5^x})$
2.) $\displaystyle 3^x(1/9)^{x+1} = 27(3^{-x}) $
Your assistance is much appreciated.
#1: I am going to use fractions instead of decimals.
$\displaystyle \frac{1}{50}=(\frac{1}{50})^{\frac{1}{2}x}$
log of both sides:
$\displaystyle log(\frac{1}{50})=\frac{1}{2}xlog(\frac{1}{50})$
$\displaystyle 1=\frac{1}{2}x$
$\displaystyle x=2$
Use your log properties.
For #2
$\displaystyle
3^x(1/9)^{x+1} = 27(3^{-x})
$
Since 9 and 27 are both powers of 3 rewrite them as their prime factorization.
$\displaystyle \frac{1}{9}=3^{-2}$ and $\displaystyle 27=3^3$
Then we get
$\displaystyle 3^x \cdot (3^{-2})^{x+1}=3^3 \cdot(3^{-x})$
Using exponent rules we get
$\displaystyle 3^x \cdot (3^{-2x-2})=(3^{-x+3})$
$\displaystyle (3^{-x-2})=(3^{-x+3})$
since the bases are the same the exponents must be equal so
$\displaystyle -x+2=-x+3$ this equation has no solutions. Therefore this equation has no solutions.