What is the answer to \sqrt[3]{(2^7-2^x)/(2^x-2)}=2 ?

The answer to \sqrt[3]{(2^7-2^x)/(2^x-2)}=2 is x=4

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\sqrt[3]{(2^7-2^x)/(2^x-2)}=2

3 \frac{2 ^{7} - 2 ^{x}}{2 ^{x} - 2} = 2

x = 4

Solution steps

3 \frac{2 ^{7} - 2 ^{x}}{2 ^{x} - 2} = 2

Rewrite the equation with

2^{x} = u

3 \frac{2 ^{7} - u}{u - 2} = 2

Solve

3 \frac{2 ^{7} - u}{u - 2} = 2 : u = 16

u = 16

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

x = 4

1 = e^{t}

1 0^{x} = 28

4^{x} = 4 + 3 \times 2^{x}

1 6^{x - 1} = 102 4^{x - 1}

- 3 e^{- 3 x} - 4 e^{- 2 x} + 5 e^{- x} + 2 = 0

What is the answer to \sqrt[3]{(2^7-2^x)/(2^x-2)}=2 ?
The answer to \sqrt[3]{(2^7-2^x)/(2^x-2)}=2 is x=4
Solve for x: \sqrt[3]{(2^7-2^x)/(2^x-2)}=2
The solution is x=4