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

The answer to log_{2}(3x-2)+log_{2}(x)=4 is x= 8/3

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

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

x = \frac{8}{3}

Decimal Notation

x = 2.66666 \dots

Solution steps

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

Apply log rules

(3 x - 2) x = 16

Solve

(3 x - 2) x = 16 : x = \frac{8}{3}, x = - 2

x = \frac{8}{3}, x = - 2

Verify Solutions:

x = \frac{8}{3}

True

, x = - 2

False

The solution is

x = \frac{8}{3}

lo g_{x + 5} (5 x) = 3

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

ln (y - 2) = 0

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

lo g_{10} (x) = 3.6274

What is the answer to log_{2}(3x-2)+log_{2}(x)=4 ?
The answer to log_{2}(3x-2)+log_{2}(x)=4 is x= 8/3
Solve for x: log_{2}(3x-2)+log_{2}(x)=4
The solution is x= 8/3