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

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

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

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

x = 3

Solution steps

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

Factor

7^{x - 1} + 7^{x - 2} + 7^{x - 3} : 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

3^{4 x^{2} - 2 x - 2} = 0

3^{5 x - 2} = 21

3^{5 x - 2} = 15

5 (2 5^{x}) - 26 (5^{x}) + 5 = 0

2 (10)^{x} + (10)^{x + 1} = 4

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