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

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Answer 1 · 2016-06-27T19:45:26

$x = \frac{1}{2}$ $x=1/2$ .

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

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

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

this equations is verified when the exponents are equal

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

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

$2 x = 1$ $2x=1$

$x = \frac{1}{2}$ $x=1/2$ .

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

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

$16^{\frac{3}{2}} = 4^{3}$ $16^(3/2)=4^3$

${(\sqrt{16})}^{3} = 4^{3}$ $(sqrt(16))^3=4^3$

here we have to be careful because the square root of $16$ $16$ can be $\pm 4$ $\pm4$ but only the solution with the $+$ $+$ is valid.

$4^{3} = 4^{3}$ $4^3=4^3$ .

How do you solve 16x+1=44x+116^(x+1)=4^(4x+1)?