Question

Solve the exponential equation

2^{3x-2} = 3^{2-x}

Collected in the board: Exponential

Steven Zheng posted 11 hours ago

Answer 1

Given

2^{3x-2} = 3^{2-x}

The next steps is to combine the variable on the same base

\dfrac{2^{3x}}{2^2} = \dfrac{3^2}{3^x}

2^{3x}\cdot 3^x = 2^2\cdot 3^2

Apply the power rule for exponet

(2^3)^x\cdot 3^x = 2^2\cdot 3^2

(2^3\cdot 3) ^x = 2^2\cdot 3^2

24^x = 36

Then we get the final answer

x = \dfrac{\log36}{\log24}

Steven Zheng posted 11 hours ago

Answer 2

Given

2^{3x-2} = 3^{2-x}

Take logarithm on both sides

\log 2^{3x-2} =\log 3^{2-x}

Apply Power Rule of Logarithm

(3x-2)\log2 = (2-x)\log3

3x\log2-2\log2 = 2\log3-x\log3

Combine the like terms

(3\log2+2\log3)x = 2\log2+3\log3

Then

x = \dfrac{2\log2+2\log3}{3\log2+\log3}

Using the power and product rules, x is simplified to

x =\dfrac{\log2^2\cdot 3^2}{\log2^3\cdot 3} =\dfrac{\log36}{\log24}

Steven Zheng posted 11 hours ago

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