What is the answer to 4^x+4^{x-1}+4^{x-2}=336 ?

The answer to 4^x+4^{x-1}+4^{x-2}=336 is x=4

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4^x+4^{x-1}+4^{x-2}=336

4^{x} + 4^{x - 1} + 4^{x - 2} = 336

x = 4

Solution steps

4^{x} + 4^{x - 1} + 4^{x - 2} = 336

Factor

4^{x} + 4^{x - 1} + 4^{x - 2} : 21 \cdot 4^{x - 2}

21 \cdot 4^{x - 2} = 336

Divide both sides by

21

4^{x - 2} = 16

Apply exponent rules

x - 2 = 2

Solve

x - 2 = 2 : x = 4

x = 4

e^{2 x} - 2 e^{x} = 0

e^{2 x} - 2 e^{x} = 3

\frac{e ^{3 x + 1}}{e ^{x - 1}} = 2

5^{x} = 21

2^{x^{2}} = 8^{2 x - 3}

What is the answer to 4^x+4^{x-1}+4^{x-2}=336 ?
The answer to 4^x+4^{x-1}+4^{x-2}=336 is x=4
Solve for x: 4^x+4^{x-1}+4^{x-2}=336
The solution is x=4