What is the answer to 7^{x-3}+7^{x-2}+7^{x-1}=57 ?

The answer to 7^{x-3}+7^{x-2}+7^{x-1}=57 is x=3

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7^{x-3}+7^{x-2}+7^{x-1}=57

7^{x - 3} + 7^{x - 2} + 7^{x - 1} = 57

x = 3

Solution steps

7^{x - 3} + 7^{x - 2} + 7^{x - 1} = 57

Factor

7^{x - 3} + 7^{x - 2} + 7^{x - 1} : 57 \cdot 7^{x - 3}

57 \cdot 7^{x - 3} = 57

Divide both sides by

57

7^{x - 3} = 1

Apply exponent rules

(x - 3) ln (7) = ln (1)

Solve

(x - 3) ln (7) = ln (1) : x = 3

x = 3

5^{x - 1} = 5^{3 x - 11}

3^{2 x - 2} = 5^{x - 1}

3 - x = 3 e^{- x}

∣ 2 (3^{x}) - 4 ∣ = 8

t, e^{- 4 u} = e^{2 t}

What is the answer to 7^{x-3}+7^{x-2}+7^{x-1}=57 ?
The answer to 7^{x-3}+7^{x-2}+7^{x-1}=57 is x=3
Solve for x: 7^{x-3}+7^{x-2}+7^{x-1}=57
The solution is x=3