# How do you solve the exponential equation 100^(7x+1)=1000^(3x-2)?

Mar 13, 2017

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

#### Explanation:

In an exponential equation: $\text{base"^"exponent}$

Make both $100$ and $1000$ be a power of $10$ so they both have the same $\text{base}$

$100 = {10}^{2}$ and $1000 = {10}^{3}$

Use the exponent rule power of a power ${\left({x}^{m}\right)}^{n} = {x}^{m \cdot n}$

${\left({10}^{2}\right)}^{7 x + 1} = {\left({10}^{3}\right)}^{3 x - 2}$

Simplify the exponents:

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

When the bases are equivalent, the exponents are equivalent:
$14 x + 2 = 9 x - 6$

Solve for $x$ by adding/subtracting like-terms:
$5 x = - 8$

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

Mar 13, 2017

I tried this:

#### Explanation:

Let us take the log in base $10$ of both sides and use properties of logs:

${\log}_{10} {100}^{7 x + 1} = {\log}_{10} {1000}^{3 x - 2}$

$\left(7 x + 1\right) {\log}_{10} 100 = \left(3 x - 2\right) {\log}_{10} 1000$

$\left(7 x + 1\right) \cdot 2 = \left(3 x - 2\right) \cdot 3$

$14 x + 2 = 9 x - 6$

$5 x = - 8$

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