What is the answer to 3^{log_{81}(2x+5)}=4 ?

The answer to 3^{log_{81}(2x+5)}=4 is x=(81^{(2ln(2))/(ln(3))}-5)/2

Solve for x: 3^{log_{81}(2x+5)}=4

The solution is x=(81^{(2ln(2))/(ln(3))}-5)/2

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3^{l o g_{81} (2 x + 5)} = 4

x = \frac{8 1 ^{\frac{2 l n ( 2 )}{l n ( 3 )}} - 5}{2}

Decimal Notation

x = 125.5

Solution steps

3^{l o g_{81} (2 x + 5)} = 4

Apply exponent rules

lo g_{81} (2 x + 5) ln (3) = ln (4)

Solve

lo g_{81} (2 x + 5) ln (3) = ln (4) : x = \frac{8 1 ^{\frac{2 l n ( 2 )}{l n ( 3 )}} - 5}{2}

x = \frac{8 1 ^{\frac{2 l n ( 2 )}{l n ( 3 )}} - 5}{2}

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

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

m, 1 = y^{6 m}

90 = 30 e^{0.03 \cdot x}

e^{2 x} - e^{x} - 56 = 0

What is the answer to 3^{log_{81}(2x+5)}=4 ?
The answer to 3^{log_{81}(2x+5)}=4 is x=(81^{(2ln(2))/(ln(3))}-5)/2
Solve for x: 3^{log_{81}(2x+5)}=4
The solution is x=(81^{(2ln(2))/(ln(3))}-5)/2