What is the answer to 5^x-97*5^{x/2}+6^4=0 ?

The answer to 5^x-97*5^{x/2}+6^4=0 is x=(8ln(3))/(ln(5)),x=(8ln(2))/(ln(5))

Solve for x: 5^x-97*5^{x/2}+6^4=0

The solution is x=(8ln(3))/(ln(5)),x=(8ln(2))/(ln(5))

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5^{x} - 97 \cdot 5^{\frac{x}{2}} + 6^{4} = 0

x = \frac{8 ln ( 3 )}{ln ( 5 )}, x = \frac{8 ln ( 2 )}{ln ( 5 )}

Decimal Notation

x = 5.46084 \dots, x = 3.44541 \dots

Solution steps

5^{x} - 97 \cdot 5^{\frac{x}{2}} + 6^{4} = 0

Apply exponent rules

5^{x} - 97 (5^{x})^{0.5} + 6^{4} = 0

Rewrite the equation with

5^{x} = u

u - 97 u^{0.5} + 6^{4} = 0

Solve

u - 97 u^{0.5} + 6^{4} = 0 : u = 6561, u = 256

u = 6561, u = 256

Substitute back

u = 5^{x},

solve for

x

Solve

5^{x} = 6561 : x = \frac{8 ln ( 3 )}{ln ( 5 )}

Solve

5^{x} = 256 : x = \frac{8 ln ( 2 )}{ln ( 5 )}

x = \frac{8 ln ( 3 )}{ln ( 5 )}, x = \frac{8 ln ( 2 )}{ln ( 5 )}

8 1^{x + 5} = 3^{2 x}

7^{3 x - 4} = 3^{x - 5}

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

0 = - e^{- x} + 3 e^{- 3 x}

2^{\frac{x}{15}} = 10

What is the answer to 5^x-97*5^{x/2}+6^4=0 ?
The answer to 5^x-97*5^{x/2}+6^4=0 is x=(8ln(3))/(ln(5)),x=(8ln(2))/(ln(5))
Solve for x: 5^x-97*5^{x/2}+6^4=0
The solution is x=(8ln(3))/(ln(5)),x=(8ln(2))/(ln(5))