# What are the solutions of 100^(7x+1) = 1000^(3x-2) ?

Mar 12, 2017

Real valued solution:

$x = - \frac{8}{5}$

Complex valued solutions:

$x = - \frac{8}{5} + \frac{2 k \pi i}{5 \ln \left(10\right)} \text{ }$ for any integer $k$

#### Explanation:

Note that if $a > 0$ then:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

So we find:

${10}^{14 x + 2} = {10}^{2 \left(7 x + 1\right)} = {\left({10}^{2}\right)}^{7 x + 1} = {100}^{7 x + 1}$

${10}^{9 x - 6} = {10}^{3 \left(3 x - 2\right)} = {\left({10}^{3}\right)}^{3 x - 2} = {1000}^{3 x - 2}$

So if:

${100}^{7 x + 1} = {1000}^{3 x - 2}$

then:

${10}^{14 x + 2} = {10}^{9 x - 6}$

Then either say "take common logarithms of both sides" or simply note that as a real-valued function ${10}^{x}$ is one to one to deduce:

$14 x + 2 = 9 x - 6$

Subtract $9 x + 2$ from both sides to get:

$5 x = - 8$

Divide both sides by $5$ to get:

$x = - \frac{8}{5}$

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Footnote

There are also Complex valued solutions, which we can deduce from Euler's identity:

${e}^{i \pi} + 1 = 0$

Hence:

${e}^{2 k \pi i} = 1 \text{ }$ for any integer $k$

Then:

$10 = {e}^{\ln 10}$

So:

${10}^{x} = {e}^{x \ln 10}$

So we find:

${10}^{\frac{2 k \pi i}{\ln} \left(10\right)} = 1 \text{ }$ for any integer $k$

Hence we find:

${10}^{14 x + 2} = {10}^{9 x - 6}$

if and only if:

$14 x + 2 = 9 x - 6 + \frac{2 k \pi i}{\ln} \left(10\right) \text{ }$ for some integer $k$

Hence:

$5 x = - 8 + \frac{2 k \pi i}{\ln} \left(10\right)$

So:

$x = - \frac{8}{5} + \frac{2 k \pi i}{5 \ln \left(10\right)}$