Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}{\log }_{x}y=3\\ {\log }_x\left(4y\right)=5\end{array}\right.$

Short answer

The solution of system of equations are $\left(2,8\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}{\log }_{x}y=3......\left(i\right)\\ {\log }_x\left(4y\right)=5......\left(ii\right)\end{array}\right.$

We know that if $y={\log }_{a}x$then $x=a^y$.

If ${\log }_{x}y=3$then role="math" localid="1646982771638" $y=x^3......\left(iii\right)$.

If ${\log }_{x}4y=5$then role="math" localid="1646982904788" $y=\frac{x^5}{4}......\left(iv\right)$.

Equate equations (iii) and (iv), then,

$$\begin{gathered} x^3=\frac{x^5}{4} \\ x=\pm 2 \end{gathered}$$

Step 2. Substitute the values of x in equation (iii).

Substitute $x=2$in equation (iii),

$$\begin{gathered} y=2^3 \\ y=8 \end{gathered}$$

Substitute $x=-2$in equation (iii),

$$\begin{gathered} y={\left(-2\right)}^3 \\ y=-8 \end{gathered}$$

Negative value is not considered as the logarithm of the negative values is not defined.

Therefore, the solution sets are$\left(2,8\right)$.