## Precalculus (10th Edition)

$x=1$.
$2^x\cdot5=10^x\\2^x\cdot5=(2\cdot5)^x\\2^x\cdot5=2^x5^x\\5=5^x\\5^1=5^x.$ We know that $a^x=a^y\longrightarrow x=y$ if $a\ne1,\ne-1$. Hence here $x=1$.