If n is a real number such that \frac{27^n + 27^n + 27^n}{3^{n+2}} = 243, what is the valu… — GRE General Test

If $n$ is a real number such that $\frac{27^n + 27^n + 27^n}{3^{n+2}} = 243$ , what is the value of $n$ ?

Answer: 3

Rewriting

27^n + 27^n + 27^n

3 \cdot (3^3)^n = 3^{3n+1}

allows the left-hand side to simplify to

\frac{3^{3n+1}}{3^{n+2}} = 3^{2n-1}

. Equating this to

243 = 3^5

gives

2n - 1 = 5

, which solves to

n = 3

Express repeated addition in the numerator as multiplication.

27^n + 27^n + 27^n = 3 \cdot 27^n

Adding three identical quantities is equivalent to multiplying one quantity by 3.

Convert base 27 to base 3 and apply exponent multiplication.

3 \cdot (3^3)^n = 3^1 \cdot 3^{3n} = 3^{3n+1}

Since

27 = 3^3

, using the power rule

(a^b)^c = a^{bc}

and product rule

a^b \cdot a^c = a^{b+c}

converts the numerator to a single power of 3.

Simplify the fraction using the quotient rule of exponents.

\frac{3^{3n+1}}{3^{n+2}} = 3^{(3n+1) - (n+2)} = 3^{2n-1}

Dividing exponential terms with the same base requires subtracting the exponent in the denominator from the exponent in the numerator.

Rewrite 243 with base 3 and equate exponents across the equal sign.

3^{2n-1} = 3^5 \implies 2n - 1 = 5

Since

243 = 3^5

, two exponential expressions with the same base are equal if and only if their exponents are equal.

Solve the linear equation for

n

2n = 6 \implies n = 3

Adding 1 to both sides yields

2n = 6

, and dividing by 2 yields

n = 3

Combining repeated addition of exponential terms and converting expressions to a common base using exponent rules.

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