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: B

Explanation:

Answer 2

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

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((a)/(b)\right)`

Therefore,

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

Simplifying,

`\log_3\left((\left(64z^2\right)^2)/(81-49)\right)`

`\log_3\left((\left(64z^2\right)^2)/(32)\right)`

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

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