Respuesta :

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given log expression

[tex]\log_2(2x^2+8x+8)[/tex]

STEP 2: Simplify the expression

[tex]\begin{gathered} \log_2(2x^2)+\log_2(8x)+\log_2(8) \\ \log_28=3 \end{gathered}[/tex]

Simplifying further,

[tex]\begin{gathered} \log _2\left(2x^2\right)+\log _2\left(8x\right) \\ \mathrm{Apply\:log\:rule}:\quad \log _a\left(x\right)+\log _a\left(y\right)=\log _a\left(xy\right) \\ \log _2\left(2x^2\right)+\log _2\left(8x\right)=\log _2\left(2x^2\cdot \:8x\right) \\ =\log _2\left(2x^2\cdot \:8x\right) \\ =\log _2\left(2^4x^3\right) \\ \mathrm{Apply\:log\:rule}:\quad \log _a\left(xy\right)=\log _a\left(x\right)+\log _a\left(y\right) \\ \log _2\left(2^4x^3\right)=\log _2\left(2^4\right)+\log _2\left(x^3\right) \\ =\log _2\left(2^4\right)+\log _2\left(x^3\right) \\ \\ \mathrm{Apply\:log\:rule}:\quad \log _a\left(a^b\right)=b \\ =4+\log _2\left(x^3\right) \end{gathered}[/tex]

Simplifying further gives:

[tex]\begin{gathered} \mathrm{Apply\:log\:rule\:}\log_a\left(x^b\right)=b\cdot\log_a\left(x\right),\:\quad\mathrm{\:assuming\:}x>0 \\ \log _2\left(x^3\right)=3\log _2\left(x\right) \\ =4+3\log _2\left(x\right) \end{gathered}[/tex]

STEP 3: Combine the derived expressions

[tex]3+4+3\log_2\left(x\right)=7+3\log_2\left(x\right)[/tex]

Hence, the answer is given as:

[tex]7+3\log _2\left(x\right)[/tex]

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