# Solve the equation for x by using base 10 logarithms. 16*4^(2.5x)=70

solve the equation using logarithms (I think this is easy level)
Solve the equation for $x$ by using base 10 logarithms.
$16\cdot {4}^{2.5x}=9$
EDIT: I made a typo (somehow... I was very far off!!)
The correct equation is this:
$16\cdot {4}^{2.5x}=70$
Can it be written like:

Then get:
${\mathrm{log}}_{10}\left(5\right)=\frac{70}{2.5x}$
The computer wants a largest value and smallest value, similar to an answer for a quadratic problem. I need to know how to get the 2 answers even if one ends up negative (I know the negative will be tossed out, but I still need to know how to get the answer).
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Matthias Calhoun
$\begin{array}{rl}16\cdot {4}^{2.5x}& =70\\ {2}^{4}\cdot \left({2}^{2}{\right)}^{2.5x}& =70\\ {2}^{4}\cdot {2}^{5x}& =70\\ {2}^{4+5x}& =70\\ {\mathrm{log}}_{10}{2}^{4+5x}& ={\mathrm{log}}_{10}70\\ \left(4+5x\right){\mathrm{log}}_{10}2& ={\mathrm{log}}_{10}70\\ 4+5x& =\frac{{\mathrm{log}}_{10}70}{{\mathrm{log}}_{10}2}\end{array}$
Can you take it from here?
$\begin{array}{rl}{4}^{2}\cdot \left({2}^{2}{\right)}^{2.5x}& =70\\ {4}^{2}\cdot \left({2}^{2}{\right)}^{2.5x}-\left(\sqrt{70}{\right)}^{2}& =0\\ \left(4\cdot {2}^{2.5x}-\sqrt{70}\right)\left(4\cdot {2}^{2.5x}+\sqrt{70}\right)& =0\end{array}$
It will yield two solutions like you want.
###### Not exactly what you’re looking for?
Harrison Mills
$16\cdot {4}^{2.5x}=16\cdot \left({4}^{2.5}{\right)}^{x}=16\cdot \left({4}^{\frac{5}{2}}{\right)}^{x}=16\cdot {32}^{x}$
So, ${32}^{x}=\frac{9}{16}$
Thus, $x={\mathrm{log}}_{32}\left(\frac{9}{16}\right)=\frac{{\mathrm{log}}_{10}\left(\frac{9}{16}\right)}{{\mathrm{log}}_{10}\left(32\right)}$
For the updated equation
$16\cdot {4}^{2.5x}=16\cdot \left({4}^{2.5}{\right)}^{x}=16\cdot \left({4}^{\frac{5}{2}}{\right)}^{x}=16\cdot {32}^{x}$
So, ${32}^{x}=\frac{30}{16}$
Thus, $x={\mathrm{log}}_{32}\left(\frac{30}{16}\right)=\frac{{\mathrm{log}}_{10}\left(\frac{15}{8}\right)}{{\mathrm{log}}_{10}\left(32\right)}$