Question

Solve the equation (linear equation) D%29%5E%7B3x%7D" id="TexFormula1" title="8^{2x+7} = (\frac{1}{32})^{3x}" alt="8^{2x+7} = (\frac{1}{32})^{3x}" align="absmiddle" class="latex-formula">

Answer 1

Answer: $x=-1$

Step-by-step explanation:

By the negative exponent rule, you have that:

$(\frac{1}{a})^n=a^{-n}$

By the exponents properties, you know that:

$(m^n)^l=m^{(nl)}$

Therefore, you can rewrite the left side of the equation has following:

$(\frac{1}{8})^{-(2x+7)}=(\frac{1}{32})^{3x}$

 Descompose 32 and 8 into its prime factors:

$32=2*2*2*2*2=2^5\\8=2*2*2=2^3$

Rewrite:

$(\frac{1}{2^3})^{-(2x+7)}=(\frac{1}{2^5})^{3x}$

Then:

$(\frac{1}{2})^{-3(2x+7)}=(\frac{1}{2})^{5(3x)}$

As the base are equal, then:

$-3(2x+7)=5(3x)$

Solve for x:

$-6x-21=15x\\-21=15x+6x\\-21=21x\\x=-1$