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

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

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log_{3}(4x^2)+log_{3}(4)=4

lo g_{3} (4 x^{2}) + lo g_{3} (4) = 4

x = \frac{9}{4}, x = - \frac{9}{4}

Decimal Notation

x = 2.25, x = - 2.25

Solution steps

lo g_{3} (4 x^{2}) + lo g_{3} (4) = 4

Move

lo g_{3} (4)

to the right side

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

Apply log rules

4 x^{2} = \frac{81}{4}

Solve

4 x^{2} = \frac{81}{4} : x = \frac{9}{4}, x = - \frac{9}{4}

x = \frac{9}{4}, x = - \frac{9}{4}

Verify Solutions:

x = \frac{9}{4}

True

, x = - \frac{9}{4}

True

The solutions are

x = \frac{9}{4}, x = - \frac{9}{4}

4 lo g_{a} (x) = 3 lo g_{10} (b)

2.39 = - lo g_{10} (x)

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

lo g_{10} (x) + 3 lo g_{10} (2) = 3

lo g_{2} (N) = 3

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