Question

If $x+{\log }_{15}(1+3^x)=x{\log }_{15}5+{\log }_{15}12$, where $x$ is an integer, then what is $x$ equal to?

• A. $-3$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

$x+{\log }_{15}(1+3^x)=x{\log }_{15}5+{\log }_{15}12$

$\Rightarrow x={\log }_{15}\left(5^x\right)+{\log }_{15}12-{\log }_{15}(1+3^x)$

$\Rightarrow x={\log }_{15}\left(\frac{5^x\times 12}{1+3^x}\right)$

$\therefore {15}^x=\frac{5^x.12}{1+3^x}{15}^x=5^x.3^x$

$\therefore 5^x.3^x=\frac{5^x.12}{1+3^x}\Rightarrow 3^x=\frac{12}{1+3x}$

$\Rightarrow 3^x+3^{2x}=12\Rightarrow x=1$

$[\log a+\log b=\log (a.b\left)\right]$

$[\log a-\log b=\log (a/b\left)\right]$.