What is the answer to log_{4}(3x)-1=log_{4-2+9}(x) ?

The answer to log_{4}(3x)-1=log_{4-2+9}(x) is x=11^{(2log_{11}(2)-log_{11}(3))/(1-2log_{11)(2)}}

Solve for x: log_{4}(3x)-1=log_{4-2+9}(x)

The solution is x=11^{(2log_{11}(2)-log_{11}(3))/(1-2log_{11)(2)}}

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lo g_{4} (3 x) - 1 = lo g_{4 - 2 + 9} (x)

x = 1 1^{\frac{2 l o g _{11} ( 2 ) - l o g _{11} ( 3 )}{1 - 2 l o g _{11} ( 2 )}}

Decimal Notation

x = 1.97767 \dots

Solution steps

lo g_{4} (3 x) - 1 = lo g_{4 - 2 + 9} (x)

Apply log rules

x = 1 1^{\frac{2 l o g _{11} ( 2 ) - l o g _{11} ( 3 )}{1 - 2 l o g _{11} ( 2 )}}

Verify Solutions:

x = 1 1^{\frac{2 l o g _{11} ( 2 ) - l o g _{11} ( 3 )}{1 - 2 l o g _{11} ( 2 )}}

True

The solution is

x = 1 1^{\frac{2 l o g _{11} ( 2 ) - l o g _{11} ( 3 )}{1 - 2 l o g _{11} ( 2 )}}

- \frac{ln ( x ) - 1}{x ^{2}} = 0

- lo g_{10} (x) = 12.3

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

lo g_{8} (\frac{x - 2}{3}) = 1

x, 6.91 = - lo g_{10} (x)

What is the answer to log_{4}(3x)-1=log_{4-2+9}(x) ?
The answer to log_{4}(3x)-1=log_{4-2+9}(x) is x=11^{(2log_{11}(2)-log_{11}(3))/(1-2log_{11)(2)}}
Solve for x: log_{4}(3x)-1=log_{4-2+9}(x)
The solution is x=11^{(2log_{11}(2)-log_{11}(3))/(1-2log_{11)(2)}}