How do you solve the exponential equation $16^(x+1)=4^(4x+1)$ ?

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Answer 1 · 2017-03-07T15:51:11

$x=1/2$

the key to these sorts of questions is to change the base numbers so that they are the same.

$16^(x+1)=4^(4x+1)$

now $16=4^2$

so $(4^2)^(x+1)=4^(4x+1)$

$4^(2x+2)=4^(4x+1)$

$=>2x+2=4x+1$

$=>2-1=4x-2x$
$1=2x$

$:.x=1/2$

How do you solve the exponential equation 16^(x+1)=4^(4x+1)16^(x+1)=4^(4x+1)?