Question

Given that p= log₂ x, find expressions in terms of p for: (a) log₂ x² A (b) log₂ (16x) (c) log8x

Answer 1

`\begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\\\ \textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}`

`p=\log_2(x) \\\\[-0.35em] ~\dotfill\\\\ \log_2(x^2) \implies 2\log_2(x)\implies \boxed{2p} \\\\[-0.35em] ~\dotfill\\\\ \log_2(16x)\implies \log_2(16)+\log_2(x)\implies \log_2(2^4)+p\implies \boxed{4+p} \\\\[-0.35em] ~\dotfill\\\\ \log_8(x)\implies \cfrac{\log_2(x)}{\log_2(8)}\implies \cfrac{p}{\log_2(2^3)}\implies \boxed{\cfrac{p}{3}}`

Answer 2

`\textsf{(a)} \quad 2p`

`\textsf{(b)} \quad 4+p`

`\textsf{(c)} \quad (p)/(3)`

Explanation:

Given:

`\boxed{p=\log_2x}`

Part (a)

`\begin{aligned}&\textsf{Given}: \quad & \log_2x^2\\&\textsf{Apply log power law}: \quad & 2 \log_2 x\\&\textsf{Substitute value of $p$}: \quad & 2p\end{aligned}`

Part (b)

`\begin{aligned}&\textsf{Given}: \quad & \log_2(16x)\\&\textsf{Apply log product law}: \quad & \log_2(16)+\log_2x\\&\textsf{Rewrite $16$ as $2^4$}: \quad & \log_2(2)^4+\log_2x\\&\textsf{Apply log power law}: \quad & 4\log_22+\log_2x\\&\textsf{Apply log law $\log_aa=1$}: \quad & 4(1)+\log_2x\\&\textsf{Simplify and substitute value of $p$}: \quad & 4+p\end{aligned}`

Part (c)

`\begin{aligned}&\textsf{Given}: \quad & \log_8x\\\\&\textsf{Apply log change of base law}: \quad & (\log_2x)/(\log_28)\\\\&\textsf{Rewrite $8$ as $2^3$}: \quad & (\log_2x)/(\log_2(2)^3)\\\\&\textsf{Apply log power law}: \quad & (\log_2x)/(3\log_2(2))\\\\&\textsf{Apply log law $\log_aa=1$}: \quad & (\log_2x)/(3(1))\\\\&\textsf{Simplify and substitute value of $p$}: \quad & (p)/(3)\\\end{aligned}`

`\boxed{\begin{minipage}{7.3 cm}\underline{Log Laws}\\\\\begin{aligned}&\textsf{Product law}: \quad & \log_axy &=\log_ax + \log_ay\\\\&\textsf{Power law}: \quad & \log_ax^n &=n\log_ax\\\\&\textsf{Log law}: \quad & \log_aa &=1\\\\&\textsf{Change of Base}: \quad &\log_ba &=(\log_xa)/(\log_xb)\\\end{aligned}\end{minipage}}`