If x+log _{15}left(5+3^xright)=x log _{15} 5+ | vedclass

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(A) Given the equation:$x+\log_{15}(5+3^x)=x\log_{15}5+\log_{15}24$Rewrite $x$ as $\log_{15}(15^x)$:$\log_{15}(15^x)+\log_{15}(5+3^x)=\log_{15}(5^x)+\

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(A) Given the equation:$x+\log_{15}(5+3^x)=x\log_{15}5+\log_{15}24$Rewrite $x$ as $\log_{15}(15^x)$:$\log_{15}(15^x)+\log_{15}(5+3^x)=\log_{15}(5^x)+\log_{15}24$Using the property $\log(a)+\log(b)=\log(ab)$:$\log_{15}(15^x(5+3^x))=\log_{15}(24 \cdot 5^x)$Equating the arguments:$15^x(5+3^x)=24 \cdot 5^x$Since $15^x = (3 \cdot 5)^x = 3^x \cdot 5^x$,we have:$3^x \cdot 5^x(5+3^x)=24 \cdot 5^x$Dividing both sides by $5^x$ (since $5^x 
eq 0$):$3^x(5+3^x)=24$Let $t=3^x$. Then:$t(5+t)=24$$t^2+5t-24=0$Factoring the quadratic equation:$(t+8)(t-3)=0$Since $t=3^x > 0$,we must have $t=3$:$3^x=3^1$$x=1$

If $x+\log _{15}\left(5+3^x\right)=x \log _{15} 5+\log _{15} 24$,then $x=\ldots .$.

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