What is the answer to 4e^{1+3x}-9e^{5-2x}=0 ?

The answer to 4e^{1+3x}-9e^{5-2x}=0 is x=(2ln(3)+4-2ln(2))/5

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4e^{1+3x}-9e^{5-2x}=0

4 e^{1 + 3 x} - 9 e^{5 - 2 x} = 0

x = \frac{2 ln ( 3 ) + 4 - 2 ln ( 2 )}{5}

Decimal Notation

x = 0.96218 \dots

Solution steps

4 e^{1 + 3 x} - 9 e^{5 - 2 x} = 0

Add

9 e^{5 - 2 x}

to both sides

4 e^{1 + 3 x} - 9 e^{5 - 2 x} + 9 e^{5 - 2 x} = 0 + 9 e^{5 - 2 x}

Simplify

4 e^{1 + 3 x} = 9 e^{5 - 2 x}

Apply exponent rules

ln (4) + 1 + 3 x = ln (9) + 5 - 2 x

Solve

ln (4) + 1 + 3 x = ln (9) + 5 - 2 x : x = \frac{2 ln ( 3 ) + 4 - 2 ln ( 2 )}{5}

x = \frac{2 ln ( 3 ) + 4 - 2 ln ( 2 )}{5}

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

4^{x} + 2^{x + 4} = 3

3^{x + 1} = 27

3^{x} = 243

1 7^{x - 2} = 4^{4 x}

What is the answer to 4e^{1+3x}-9e^{5-2x}=0 ?
The answer to 4e^{1+3x}-9e^{5-2x}=0 is x=(2ln(3)+4-2ln(2))/5