Chapter 3.3, Problem 75E

To determine

To condense the expression $\frac{1}{3}\left[{\log }_{8}y+2{\log }_8\left(y+4\right)\right]-{\log }_8\left(y-1\right)$ to the logarithm of a single quantity.

Answer to Problem 75E

  $\frac{1}{3}\left[{\log }_{8}y+2{\log }_8\left(y+4\right)\right]-{\log }_8\left(y-1\right)={\log }_8\frac{{\left(y{\left(y+4\right)}^2\right)}^{\frac{1}{3}}}{\left(y-1\right)}$

Explanation of Solution

Given information:

Consider the expression $\frac{1}{3}\left[{\log }_{8}y+2{\log }_8\left(y+4\right)\right]-{\log }_8\left(y-1\right)$

Calculations:

First, apply power property for Logarithm with base $a\ne 1$ , ${\log }_a{u}^n=n{\log }_{a}u$ where n is a real number and u is a positive real number.

  $\Rightarrow \frac{1}{3}\left[{\log }_{8}y+{\log }_8{\left(y+4\right)}^2\right]-{\log }_8\left(y-1\right)$

Now use Product property for Logarithm with base $a\ne 1$ , that is., ${\log }_a\left(uv\right)={\log }_{a}u+{\log }_{a}v$ where u and v are positive real numbers.

  $\Rightarrow \frac{1}{3}\left[{\log }_8\left(y{\left(y+4\right)}^2\right)\right]-{\log }_8\left(y-1\right)$

Again, apply power property and get the final result:

  $\Rightarrow {\log }_8{\left(y{\left(y+4\right)}^2\right)}^{\frac{1}{3}}-{\log }_8\left(y-1\right)$

Use Quotient property forLogarithm with base $a\ne 1$ , that is., ${\log }_a\frac{u}{v}={\log }_{a}u-{\log }_{a}v$ where u and v are positive real numbers.

  $\Rightarrow {\log }_8\frac{{\left(y{\left(y+4\right)}^2\right)}^{\frac{1}{3}}}{\left(y-1\right)}$

Hence, condensed form of the expression: $\frac{1}{3}\left[{\log }_{8}y+2{\log }_8\left(y+4\right)\right]-{\log }_8\left(y-1\right)={\log }_8\frac{{\left(y{\left(y+4\right)}^2\right)}^{\frac{1}{3}}}{\left(y-1\right)}$