Question

Solve the logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answer. log6x2 = log6(5x + 36)

Answer 1

`\bf log_6(x^2)=log_6(5x+36)\impliedby \textit{removing the logs} \\\\\\ x^2=5x+36\implies \begin{array}{lcclll} x^2&-5x&-36&=0\\ &\uparrow &\uparrow \\ &-9+4&-9\cdot 4 \end{array} \\\\\\ (x-9)(x+4)=0\implies x= \begin{cases} 9\\ -4 \end{cases}`

domain for a logarithm is that, the value of the expression has to be positive, logarithms will never use a negative value, so in this case, both cases provide a positive value

Answer 2

Answer:x=-4 or 9

Explanation:

Given

`log_(6)( x^2)=log_(6)(5x+36)`

This log will be defined when

5x+36>0

`x>-(36)/(5)`

L.H.S=R.H.S

thus
`x^2=5x+36`

`x^2-5x-36=0`

`x^2-9x+4x-36=0`

`\left ( x+4\right )\left ( x-9\right )=0`

thus x=-4 or 9

as these two values are in the domain therefore -4 & 9 are the solution of the Given system.