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Popular >

2ln(2x)=1+ln(a)

Solution

2 ln (2 x) = 1 + ln (a)

Solution

x = \frac{e ^{\frac{1 + l n ( a )}{2}}}{2} {a > 0}

Solution steps

2 ln (2 x) = 1 + ln (a)

Divide both sides by

2

ln (2 x) = \frac{1 + ln ( a )}{2}

Apply log rules

2 x = e^{\frac{1 + l n ( a )}{2}}

Solve

2 x = e^{\frac{1 + l n ( a )}{2}} : x = \frac{e ^{\frac{1 + l n ( a )}{2}}}{2}

x = \frac{e ^{\frac{1 + l n ( a )}{2}}}{2}

Verify Solutions:

x = \frac{e ^{\frac{1 + l n ( a )}{2}}}{2} {a > 0}

The solution is

x = \frac{e ^{\frac{1 + l n ( a )}{2}}}{2} {a > 0}

Popular Examples

log_{10}(4)+2log_{10}(x)=6

lo g_{10} (4) + 2 lo g_{10} (x) = 6

log_{5}(y)=4-4log_{y}(5)

lo g_{5} (y) = 4 - 4 lo g_{y} (5)

log_{10}(x+4)+log_{10}(x)=9

lo g_{10} (x + 4) + lo g_{10} (x) = 9

4 ln (2 x + 9) + 5 = 29

log_{x}(9x-20)=2

lo g_{x} (9 x - 20) = 2

Frequently Asked Questions (FAQ)

What is the answer to 2ln(2x)=1+ln(a) ?
The answer to 2ln(2x)=1+ln(a) is x=(e^{(1+ln(a))/2})/(2) {a>0}