Question

Find the value of x in log(x+3) +log(x-3) =log(72)​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★$log_{10}^{x+3}+log_{10}^{x-3}=log_{10}^{72}$

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The value of x.

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We would use the formulas related to logarithms.

$\huge\tt{\red{\underline{Answer:}}}$

We have,

$\implies log_{10}^{x+3}+log_{10}^{x-3}=log_{10}^{72}$

$\implies log_{10}^{(x+3) (x-3) }=log_{10}^{72}$

$\large\green{\boxed{log_{a}^{b}+log_{a}^{c}=log_{a}^{bc}}}$

$\implies log_{10}^{x^{2}-3^{2}}=log_{10}^{72}$

$\large\purple{\boxed{a^{2}-b^{2}=(a+b)(a-b)}}$

$\implies log_{10}^{x^{2}-9}=log_{10}^{72}$

Eliminating log on both sides, we have:

$\implies\cancel{ log_{10}}^{x^{2}-9}=\cancel{log_{10}}^{72}$

$\implies x^{2}-9=72$

$\implies x^{2}=(72+9)$

$\implies x^{2}=81$

$\implies x^{2}=9^{2}$

${\underline{\boxed{\red{x=9}}}}$

Therefore the value of x is 9.