Question

Solve the exponential equation by rewriting the base. explain the steps towards your answer.

Answer 1

Answer:

n=0

Step-by-step explanation:

Given in the question an equation

$\frac{1}{9}^{2n}=27^{n}$

we can write

$\frac{1}{9}$

as

$(\frac{1}{3})^{2}$

further as

$3^{-2}$

so

$3^{-2(2n)}=3^{3n}$

$3^{-4n}=3^{3n}$

Since now the base is same

and we know that

$x^{n}=x^{m}\\n=m$

-4n = 3n

n cancel out which means n =0

Answer 2

Answer: $n=0$

Step-by-step explanation:

You need to remember that:

$a^n=a^m\\n=m$

By the Power of a power property:

$(a^m)^n=a^{mn}$

By the Negative exponent rule:

$a^{-1}=\frac{1}{a}$

Therefore, having the expression $(\frac{1}{9} )^{2n}=27^{n}$:

Descompose 27 and 9 into their prime factors:

$27=3*3*3=3^3$

$9=3*3=3^2$

Rewrite the expression and simplify. Then:

$(3^{-2})^{2n}=(3^3)^n\\3^{-4n}=3^{3n}\\-4n=3n\\-4n-3n=0\\n=0$