What is the answer to 25^x-20^x-16^x=0 ?

The answer to 25^x-20^x-16^x=0 is x=(ln((1+sqrt(5))/2))/(ln(5/4))

Solve for x: 25^x-20^x-16^x=0

The solution is x=(ln((1+sqrt(5))/2))/(ln(5/4))

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2 5^{x} - 2 0^{x} - 1 6^{x} = 0

x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{5}{4} )}

Decimal Notation

x = 2.15651 \dots

Solution steps

2 5^{x} - 2 0^{x} - 1 6^{x} = 0

Divide both sides by

1 6^{x}

\frac{2 5 ^{x}}{1 6 ^{x}} - \frac{2 0 ^{x}}{1 6 ^{x}} - \frac{1 6 ^{x}}{1 6 ^{x}} = \frac{0}{1 6 ^{x}}

Simplify

(\frac{25}{16})^{x} - (\frac{5}{4})^{x} - 1 = 0

Apply exponent rules

((\frac{5}{4})^{x})^{2} - (\frac{5}{4})^{x} - 1 = 0

Rewrite the equation with

(\frac{5}{4})^{x} = u

(u)^{2} - u - 1 = 0

Solve

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

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

Substitute back

u = (\frac{5}{4})^{x},

solve for

x

Solve

(\frac{5}{4})^{x} = \frac{1 + 5}{2} : x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{5}{4} )}

Solve

(\frac{5}{4})^{x} = \frac{1 - 5}{2} :

No Solution for

x \in R

x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{5}{4} )}

3^{x + 5} = \frac{3 ^{x + 3} + 8}{9}

x, 8^{2 x} \cdot 3^{- \frac{1}{2}} \cdot y = 36

5^{x} + 36 x - 97 = 0

(x + 2) e^{x - 1} - 1 = 0

2^{3^{8^{x}}} = 512

What is the answer to 25^x-20^x-16^x=0 ?
The answer to 25^x-20^x-16^x=0 is x=(ln((1+sqrt(5))/2))/(ln(5/4))
Solve for x: 25^x-20^x-16^x=0
The solution is x=(ln((1+sqrt(5))/2))/(ln(5/4))