What is the answer to log_{7}(3^{2x}-3x+1)=1 ?

The answer to log_{7}(3^{2x}-3x+1)=1 is x=1,x=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{2ln(3)}

Solve for x: log_{7}(3^{2x}-3x+1)=1

The solution is x=1,x=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{2ln(3)}

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lo g_{7} (3^{2 x} - 3 x + 1) = 1

x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

Decimal

x = 1, x = - 1.99584 \dots

Solution steps

lo g_{7} (3^{2 x} - 3 x + 1) = 1

Apply log rules

9^{x} - 3 x + 1 = 7

Solve

9^{x} - 3 x + 1 = 7 : x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

lo g_{49} (x - 6) = 0.5

2 = lo g_{x} (9)

lo g_{10} (\frac{2}{x}) = - 2

lo g_{3} (2 x - 1) = 4

ln (x^{2} - 63) = 0

What is the answer to log_{7}(3^{2x}-3x+1)=1 ?
The answer to log_{7}(3^{2x}-3x+1)=1 is x=1,x=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{2ln(3)}
Solve for x: log_{7}(3^{2x}-3x+1)=1
The solution is x=1,x=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{2ln(3)}