What is the answer to 9^x-6^x=4^x ?

The answer to 9^x-6^x=4^x is x=(ln((1+sqrt(5))/2))/(ln(3/2))

Solve for x: 9^x-6^x=4^x

The solution is x=(ln((1+sqrt(5))/2))/(ln(3/2))

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9^{x} - 6^{x} = 4^{x}

x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{3}{2} )}

Decimal

x = 1.18681 \dots

Solution steps

9^{x} - 6^{x} = 4^{x}

Divide both sides by

4^{x}

\frac{9 ^{x}}{4 ^{x}} - \frac{6 ^{x}}{4 ^{x}} = \frac{4 ^{x}}{4 ^{x}}

Simplify

(\frac{9}{4})^{x} - (\frac{3}{2})^{x} = 1

Apply exponent rules

((\frac{3}{2})^{x})^{2} - (\frac{3}{2})^{x} = 1

Rewrite the equation with

(\frac{3}{2})^{x} = u

(u)^{2} - u = 1

Solve

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

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

Substitute back

u = (\frac{3}{2})^{x},

solve for

x

Solve

(\frac{3}{2})^{x} = \frac{1 + 5}{2} : x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{3}{2} )}

Solve

(\frac{3}{2})^{x} = \frac{1 - 5}{2} :

No Solution for

x \in R

x = \frac{ln ( \frac{1 + 5}{2} )}{ln ( \frac{3}{2} )}

4^{x} + 2^{1 + 2 x} = 50

1 0^{- 2 x + 3} = 100

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

4 + 7 e^{- 2 x} = 3 e^{2 x}

x, e^{\frac{x}{y}} = x - y

What is the answer to 9^x-6^x=4^x ?
The answer to 9^x-6^x=4^x is x=(ln((1+sqrt(5))/2))/(ln(3/2))
Solve for x: 9^x-6^x=4^x
The solution is x=(ln((1+sqrt(5))/2))/(ln(3/2))