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

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Answer 1 · 2016-06-27T19:45:26

$x=1/2$ .

You have that $16=4^2$ , you can write

$16^(x+1)=4^(2(x+1))$ then we have

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

this equations is verified when the exponents are equal

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

$2x+2=4x+1$

$2x=1$

$x=1/2$ .

We can verify if this solution is correct substituting in the original equation

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

$16^(3/2)=4^3$

$(sqrt(16))^3=4^3$

here we have to be careful because the square root of $16$ can be $\pm4$ but only the solution with the $+$ is valid.

$4^3=4^3$ .

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