Question

2log(x+3)=log81 find x

Answer 1

$\textbf{Given:}$

$2\,\log(x+3)=\log\,81$

$\textbf{To find:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\text{Consider,}$

$2\,\log(x+3)=\log\,81$

$\text{Using power rule of logarithm}$

$\boxed{\bf\,log_aM^n=n\;log_aM}$

$\log(x+3)^2=\log\,81$

$(x+3)^2=81$

$x^2+6x+9=81$

$x^2+6x-72=0$

$(x+12)(x-6)=0$

$\implies\,x=6,-12$

$\therefore\textbf{The values of x are 6 and -12}$

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Answer 2

Answer:

2logx+2log3=log 9^2

logx^2+log3^2=log81

logx^2+log9=log81

logx^2=log81-log9

logx^2=log(81/9)

logx^2=log9

x^2=9

x=+3or-3

Step-by-step explanation:

since logm-logn is logm by n