# How do you solve 2^x*5=10^x?

Jun 30, 2016

If $A = B$, then $\log A = \log B$:

$\log \left({2}^{x} \times 5\right) = \log \left({10}^{x}\right)$

Use the rule ${\log}_{a} \left(\frac{n}{m}\right) = {\log}_{a} n - {\log}_{a} m$ to simplify.

$\log {2}^{x} + \log 5 = \log {10}^{x}$

Use the rule $\log {a}^{n} = n \log a$ to simplify further.

$x \log 2 + \log 5 = x \log 10$

$\log 5 = x \log 10 - x \log 2$

Factor out the $x$.

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

Put back the logs on the left side into quotient form to simplify.

$\log 5 = x \left(\log \left(\frac{10}{2}\right)\right)$

$\log 5 = x \left(\log 5\right)$

Isolate x.

$\log \frac{5}{\log} 5 = x$

$1 = x$

Checking in the original equation, we find that this solution works.

Hopefully this helps!

Jun 30, 2016

$x = 1$ (using alternate solution method)

#### Explanation:

If ${2}^{x} \cdot 5 = {10}^{x}$

$\Rightarrow {2}^{x} \cdot 5 = {\left(2 \cdot 5\right)}^{x}$

$\Rightarrow {2}^{x} \cdot 5 = {2}^{x} \cdot {5}^{x}$

$\Rightarrow 5 = {5}^{x}$

$\Rightarrow x = 1$