How do I solve this exponential equation?

Question

I have been trying to use substitution but that doesn't work unless its $4 x^{2}$ $4x^2$ .

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Answer 1 · 2018-06-27T11:29:28

$x = 4, or, x = 3$ $x=4, or, x=3$ .

Let, $2^{x} = y . Then, 4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2} = y^{2}$ $2^x=y." Then, "4^x=(2^2)^x=(2^x)^2=y^2$ .

Also, $2^{x + 3} = 2^{x} \cdot 2^{3} = 8 y$ $2^(x+3)=2^x*2^3=8y$ .

Thus, the eqn. becomes, $y^{2} - 3 \cdot 8 y + 128 = 0,$ $y^2-3*8y+128=0,$

$i . e ., y^{2} - 24 y + 128 = 0$ $i.e., y^2-24y+128=0$ .

$:. (y-16)(y-8)=0$ .

$:. y=16, or, y=8$

$:. 2^x=16=2^4, or, y=8=2^3$ .

$:. x=4, or, x=3$ .