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# Solve for \$x\$: \$\$\dfrac{66-2^x}{2^x+3}=\dfrac{4-2^x}{2^{x+1}+6}\$\$

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Solve for \$x\$: \(\dfrac{66-2^x}{2^x+3}=\dfrac{4-2^x}{2^{x+1}+6}\)

michaelcai  Sep 17, 2017
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Solve for x over the real numbers:
(66 - 2^x)/(2^x + 3) = (4 - 2^x)/(2^(x + 1) + 6)

Hint: | Expand out terms of the left hand side.
(66 - 2^x)/(2^x + 3) = 66/(2^x + 3) - 2^x/(2^x + 3):
66/(2^x + 3) - 2^x/(2^x + 3) = (4 - 2^x)/(2^(x + 1) + 6)

Hint: | Look for an expression to multiply both sides by in order to clear fractions.
Multiply both sides by 2^(x + 1) + 6:
132 - 2^(x + 1) = 4 - 2^x

Hint: | Move everything to the left hand side.
Subtract 4 - 2^x from both sides:
128 + 2^x - 2^(x + 1) = 0

Hint: | Simplify 128 + 2^x - 2^(x + 1) = 0 by making a substitution.
Simplify and substitute y = 2^x.
128 + 2^x - 2^(x + 1) = 128 - 2^x
= 128 - y:
128 - y = 0

Hint: | Divide both sides by the sign of the leading coefficient of 128 - y.
Multiply both sides by -1:
y - 128 = 0

Hint: | Solve for y.
y = 128

Hint: | Perform back substitution on y = 128.
Substitute back for y = 2^x:
2^x = 128

Hint: | Perform the prime factorization of the right hand side.
128 = 2^7:
2^x = 2^7

Hint: | Equate exponents.
Equate exponents of 2 on both sides:
x = 7

Guest Sep 17, 2017

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