What is the answer to 49^x-6*7^x+5=0 ?

The answer to 49^x-6*7^x+5=0 is x=(ln(5))/(ln(7)),x=0

Solve for x: 49^x-6*7^x+5=0

The solution is x=(ln(5))/(ln(7)),x=0

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4 9^{x} - 6 \cdot 7^{x} + 5 = 0

x = \frac{ln ( 5 )}{ln ( 7 )}, x = 0

Decimal Notation

x = 0.82708 \dots, x = 0

Solution steps

4 9^{x} - 6 \cdot 7^{x} + 5 = 0

Apply exponent rules

(7^{x})^{2} - 6 \cdot 7^{x} + 5 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} - 6 u + 5 = 0

Solve

u^{2} - 6 u + 5 = 0 : u = 5, u = 1

u = 5, u = 1

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 5 : x = \frac{ln ( 5 )}{ln ( 7 )}

Solve

7^{x} = 1 : x = 0

x = \frac{ln ( 5 )}{ln ( 7 )}, x = 0

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

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

\frac{2 ^{3 x - 1}}{2 ^{x + 5}} = 2^{x}

2^{x + 14} = 1 6^{2 x}

8^{- 4} = 4^{- 2 x - 4}

What is the answer to 49^x-6*7^x+5=0 ?
The answer to 49^x-6*7^x+5=0 is x=(ln(5))/(ln(7)),x=0
Solve for x: 49^x-6*7^x+5=0
The solution is x=(ln(5))/(ln(7)),x=0