Please can someone take me through the steps in this as I have the x's on both sides
$\displaystyle 7^(x-3)=4^(2x)$
The (x-3) and (2x) are exponents
My math type didnt import the text in
Thank you
$\displaystyle 7^{x-3} = 4^{2x}$
$\displaystyle \log\left(7^{x-3}\right) = \log\left(4^{2x}\right)
$
$\displaystyle (x-3)\log{7} = 2x\log{4}$
$\displaystyle x\log{7} - 3\log{7} = 2x\log{4}$
$\displaystyle x\log{7} - 2x\log{4} = 3\log{7}
$
$\displaystyle x(\log{7} - 2\log{4}) = 3\log{7}$
$\displaystyle x = \frac{3\log{7}}{\log{7} - 2\log{4}}$
in future, start a new problem with a new thread.
using change of base, note that
$\displaystyle \log_x(5^9) = \frac{\log_5(5^9)}{\log_5{x}} = \frac{9}{\log_5{x}}$
now you have the equation ...
$\displaystyle \frac{9}{\log_5{x}} = \log_5{x}$
$\displaystyle 9 = (\log_5{x})^2$
$\displaystyle \pm 3 = \log_5{x}$
$\displaystyle x = 5^3$ or $\displaystyle x=5^{-3}$