Respuesta :
[tex]\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}[/tex]
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
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:x=-4 or 9
Step-by-step explanation:
Given
[tex]log_{6}( x^2)=log_{6}(5x+36)[/tex]
This log will be defined when
5x+36>0
[tex]x>-\frac{36}{5}[/tex]
L.H.S=R.H.S
thus [tex]x^2=5x+36[/tex]
[tex]x^2-5x-36=0[/tex]
[tex]x^2-9x+4x-36=0[/tex]
[tex]\left ( x+4\right )\left ( x-9\right )=0[/tex]
thus x=-4 or 9
as these two values are in the domain therefore -4 & 9 are the solution of the Given system.
