What is the answer to 4ln(4b)-ln(2b)=0 ?

The answer to 4ln(4b)-ln(2b)=0 is b= 1/(e^{\frac{7ln(2)){3}}}

Solve for b: 4ln(4b)-ln(2b)=0

The solution is b= 1/(e^{\frac{7ln(2)){3}}}

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4 ln (4 b) - ln (2 b) = 0

b = \frac{1}{e ^{\frac{7 l n ( 2 )}{3}}}

Solution steps

4 ln (4 b) - ln (2 b) = 0

Apply log rules

b = \frac{1}{e ^{\frac{7 l n ( 2 )}{3}}}

Verify Solutions:

b = \frac{1}{e ^{\frac{7 l n ( 2 )}{3}}}

True

The solution is

b = \frac{1}{e ^{\frac{7 l n ( 2 )}{3}}}

- 0.3 = - 0.56585 - 0.059 ln (x)

lo g_{10} (36) + lo g_{10} (x) = 25

lo g_{2} (x + 2) = 35

lo g_{\frac{1}{2}} (2 x - 3) = 0

lo g_{3} (2 x + 3) = 5 - lo g_{3} (x + 6)

What is the answer to 4ln(4b)-ln(2b)=0 ?
The answer to 4ln(4b)-ln(2b)=0 is b= 1/(e^{\frac{7ln(2)){3}}}
Solve for b: 4ln(4b)-ln(2b)=0
The solution is b= 1/(e^{\frac{7ln(2)){3}}}