Alg9.4.34

If $\frac{\log(\sqrt{x+1}+1)}{\log(\sqrt[3]{x-40})}=3\,$ Find the value of x

Simplifying

$\log(\sqrt{x+1}+1)=3\log(\sqrt[3]{x-40})\,$

$\log(\sqrt{x+1}+1)=\log(x-40)\,$

Removing logs bothsides,

$\sqrt{x+1}+1=x-40\,$

$\sqrt{x+1}=x-41\,$

Squaring on both sides,

$x+1=x^2+1681-82x\,$

$x^2-83x+1680=0\,$

$(x-48)(x-35)=0\,$

The values of x are 48 and 35.