What is the answer to (log_{10}(6-x^2))/(log_{10)(3x-4)}=2 ?

The answer to (log_{10}(6-x^2))/(log_{10)(3x-4)}=2 is x=(6+sqrt(11))/5

Solve for x: (log_{10}(6-x^2))/(log_{10)(3x-4)}=2

The solution is x=(6+sqrt(11))/5

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\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = 2

x = \frac{6 + 11}{5}

Decimal Notation

x = 1.86332 \dots

Solution steps

\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = 2

Multiply both sides by

lo g_{10} (3 x - 4)

\frac{lo g _{10} ( 6 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} lo g_{10} (3 x - 4) = 2 lo g_{10} (3 x - 4)

Simplify

lo g_{10} (6 - x^{2}) = 2 lo g_{10} (3 x - 4)

Apply log rules

6 - x^{2} = (3 x - 4)^{2}

Solve

6 - x^{2} = (3 x - 4)^{2} : x = \frac{6 + 11}{5}, x = \frac{6 - 11}{5}

x = \frac{6 + 11}{5}, x = \frac{6 - 11}{5}

Verify Solutions:

x = \frac{6 + 11}{5}

True

, x = \frac{6 - 11}{5}

False

The solution is

x = \frac{6 + 11}{5}

lo g_{4} (x + 12) - lo g_{4} (5 - x) = 2

lo g_{2} (x + 7) = 5

lo g_{b} (3) = \frac{1}{2}

ln ((2 x + 3)^{3}) = 1

\frac{lo g _{10} ( 2 x ^{2} + 1 )}{lo g _{10} ( x + 1 )} = 2

What is the answer to (log_{10}(6-x^2))/(log_{10)(3x-4)}=2 ?
The answer to (log_{10}(6-x^2))/(log_{10)(3x-4)}=2 is x=(6+sqrt(11))/5
Solve for x: (log_{10}(6-x^2))/(log_{10)(3x-4)}=2
The solution is x=(6+sqrt(11))/5