Question

What is the value of \(7^{\left(3+\log_7 5\right)} \) ?

• A. 1575
• B. 3140
• C. 1715
• D. 3460

Hint:

Split $7^{3+\log_7 5}$ into two powers with base $7$. One factor simplifies immediately by the inverse exponent-logarithm identity.

Answer 1

The Correct Option is A

Step 1: Split the exponent using laws of exponents.
\[ 7^{\left(3+\log_7 5\right)} = 7^3 \cdot 7^{\log_7 5} \] Step 2: Apply logarithmic identity.
Using \(a^{\log_a b} = b\):
\[ 7^{\log_7 5} = 5 \] Step 3: Multiply the terms.
\[ 7^3 = 343 \]
\[ 343 \times 5 = 1715 \] Step 4: Final calculation.
\[ 7^{\left(3+\log_7 5\right)} = 1715 \] Final Answer:
\[ \boxed{1715} \]

Answer 2

Concept:

• A sum in an exponent becomes a product of powers.
• The identity $a^{\log_a x}=x$ removes the logarithmic exponent directly.

Step 1: Separate the two exponent terms.
$7^{3+\log_7 5}=7^3\cdot7^{\log_7 5}$.

Step 2: Simplify each factor.
$7^3=343$ and $7^{\log_7 5}=5$.

Step 3: Multiply the factors.
$343\times5=1715$.

Final Answer: $1715$, option C