What is the answer to 3^{x-1}+3^{4-x}=36 ?

The answer to 3^{x-1}+3^{4-x}=36 is x=(ln(9(6+sqrt(33))))/(ln(3)),x=(ln(9(6-sqrt(33))))/(ln(3))

Solve for x: 3^{x-1}+3^{4-x}=36

The solution is x=(ln(9(6+sqrt(33))))/(ln(3)),x=(ln(9(6-sqrt(33))))/(ln(3))

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3^{x - 1} + 3^{4 - x} = 36

x = \frac{ln ( 9 ( 6 + 33 ) )}{ln ( 3 )}, x = \frac{ln ( 9 ( 6 - 33 ) )}{ln ( 3 )}

Decimal Notation

x = 4.24227 \dots, x = 0.75772 \dots

Solution steps

3^{x - 1} + 3^{4 - x} = 36

Apply exponent rules

3^{x} \cdot 3^{- 1} + (3^{x})^{- 1} \cdot 3^{4} = 36

Rewrite the equation with

3^{x} = u

u \cdot 3^{- 1} + (u)^{- 1} \cdot 3^{4} = 36

Solve

u \cdot 3^{- 1} + u^{- 1} \cdot 3^{4} = 36 : u = 9 (6 + 33), u = 9 (6 - 33)

u = 9 (6 + 33), u = 9 (6 - 33)

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 9 (6 + 33) : x = \frac{ln ( 9 ( 6 + 33 ) )}{ln ( 3 )}

Solve

3^{x} = 9 (6 - 33) : x = \frac{ln ( 9 ( 6 - 33 ) )}{ln ( 3 )}

x = \frac{ln ( 9 ( 6 + 33 ) )}{ln ( 3 )}, x = \frac{ln ( 9 ( 6 - 33 ) )}{ln ( 3 )}

ln (e^{x} - 3) = 2

5^{2 x} = 20

4 - 3^{x} = 0

1 6^{2 x + 1} = 9^{3 x - 4}

1 6^{2 x - 3} 4^{- 2 x} = 16

What is the answer to 3^{x-1}+3^{4-x}=36 ?
The answer to 3^{x-1}+3^{4-x}=36 is x=(ln(9(6+sqrt(33))))/(ln(3)),x=(ln(9(6-sqrt(33))))/(ln(3))
Solve for x: 3^{x-1}+3^{4-x}=36
The solution is x=(ln(9(6+sqrt(33))))/(ln(3)),x=(ln(9(6-sqrt(33))))/(ln(3))