# How do you solve 4^x * 5^(4x+3) = 10^(2x+3)?

Mar 28, 2016

$x \approx 0.65$

#### Explanation:

$1$. Start by taking the logarithm of both sides, since the bases are not the same on the left and right sides of the equation.

${4}^{x} \cdot {5}^{4 x + 3} = {10}^{2 x + 3}$

$\log \left({4}^{x} \cdot {5}^{4 x + 3}\right) = \log \left({10}^{2 x + 3}\right)$

$2$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m} \cdot \textcolor{b l u e}{n}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) + {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the left side of the equation.

$\log \left({4}^{x}\right) + \log \left({5}^{4 x + 3}\right) = \log \left({10}^{2 x + 3}\right)$

$3$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to rewrite both sides of the equation.

$x \log \left(4\right) + \left(4 x + 3\right) \log \left(5\right) = \left(2 x + 3\right) \log \left(10\right)$

$4$. Expand the brackets.

$x \log \left(4\right) + 4 x \log \left(5\right) + 3 \log \left(5\right) = 2 x \log \left(10\right) + 3 \log \left(10\right)$

$5$. Bring all terms with the variable, $x$, to the left side of the equation and all terms without to the right side.

$x \log \left(4\right) + 4 x \log \left(5\right) - 2 x \log \left(10\right) = 3 \log \left(10\right) - 3 \log \left(5\right)$

$6$. Factor out $x$ from the terms on the left side.

$x \left(\log \left(4\right) + 4 \log \left(5\right) - 2 \log \left(10\right)\right) = 3 \log \left(10\right) - 3 \log \left(5\right)$

$7$. Solve for $x$.

$x = \frac{3 \log \left(10\right) - 3 \log \left(5\right)}{\log \left(4\right) + 4 \log \left(5\right) - 2 \log \left(10\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 0.65 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Apr 16, 2016

$0.646$

#### Explanation:

${4}^{x} \cdot {5}^{4 x + 3} = {10}^{2 x + 3}$

$\implies {4}^{x} \cdot {5}^{4 x} \cdot {5}^{3} = {10}^{2 x} \cdot {10}^{3}$

$\implies {\left(4 \cdot {5}^{4}\right)}^{x} \cdot {5}^{3} = {\left({10}^{2}\right)}^{x} \cdot {10}^{3}$

Dividing both sides by ${5}^{3} \cdot {\left({10}^{2}\right)}^{x}$

$\implies {\left(\frac{4 \cdot {5}^{4}}{10} ^ 2\right)}^{x} = {10}^{3} / {5}^{3}$

$\implies {\left(\frac{\cancel{4} \cdot \cancel{5} \times \cancel{5} \times {5}^{2}}{\cancel{10} \times \cancel{10}}\right)}^{x} = \frac{1000}{125} = 8 = {2}^{3}$

Taking${\log}_{10}$ on both sides and using property $\log {m}^{n} = n \log m$ we get
$\implies {\log}_{10} {5}^{2 x} = {\log}_{10} {2}^{3}$
$\implies 2 x \cdot {\log}_{10} 5 = 3 \cdot {\log}_{10} 2$

Dividing both sides by $\left(2 \cdot {\log}_{10} 5\right)$

$\implies x = \frac{3 \cdot {\log}_{10} 2}{2 \cdot {\log}_{10} 5} = 0.646$