Question

What is 2 (log Subscript 3 Baseline 8 + log Subscript 3 Baseline z) minus log Subscript 3 Baseline (3 Superscript 4 Baseline minus 7 squared) written as a single logarithm?

Answer 1

Answer:

Value of expression in single logarithm is $\log_3\left(2z^2\right)$.

Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

$\log_3\left(\left(64z^2\right)\right)-\log_3\left(3^4-7^2\right)$

Applying quotient rule of logarithmic,

$\log_c\left(a\right)-\log_c\left(b\right)=\log_c\left(\frac{a}{b}\right)$

Therefore,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{3^4-7^2}\right)$

Simplifying,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{81-49}\right)$

$\log_3\left(\dfrac{\left(64z^2\right)^2}{32}\right)$

$\log_3\left(2z^2\right)$

Therefore value of expression is $\log_3\left(2z^2\right)$

Answer 2

Answer: B

Step-by-step explanation: