How do you solve $4^x - 2^(x+1)-15=0$ ?

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Answer 1 · 2018-04-18T18:30:05

$x=log_2 5~~2.322$ #.

Observe that, $4^x=(2^2)^x=(2^x)^2, and, 2^(x+1)=2^x*2^1$ .

Therefore, the given eqn. is, $(2^x)^2-2*2^x-15=0$ .

So, if we subst. $2^x=y$ , the eqn. becomes,

$y^2-2y-15=0$ .

$(y-5)(y+3)=0$ .

$:. y=5, or, y=-3$ .

Since, $y=2^x gt 0, y=-3$ is not admissible.

$:. y=5 rArr 2^x=5$ .

$:. x=log_2 5$ .

Using the Change of Base Rule, $x=log_10 5/log_10 2$ ,

$=0.699/0.301~~2.322$

How do you solve 4^x - 2^(x+1)-15=04^x - 2^(x+1)-15=0?