What is the answer to log_{2}(4-2^x)=0 ?

The answer to log_{2}(4-2^x)=0 is x=(ln(3))/(ln(2))

Solve for x: log_{2}(4-2^x)=0

The solution is x=(ln(3))/(ln(2))

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lo g_{2} (4 - 2^{x}) = 0

x = \frac{ln ( 3 )}{ln ( 2 )}

Decimal Notation

x = 1.58496 \dots

Solution steps

lo g_{2} (4 - 2^{x}) = 0

Apply log rules

4 - 2^{x} = 1

Solve

4 - 2^{x} = 1 : x = \frac{ln ( 3 )}{ln ( 2 )}

x = \frac{ln ( 3 )}{ln ( 2 )}

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

- lo g_{10} (x) = 5.2

ln (x - 1) = 2

ln (2) + ln (x - 3) = ln (x)

ln (x - 2) + ln (x + 5) = 2 ln (3)

What is the answer to log_{2}(4-2^x)=0 ?
The answer to log_{2}(4-2^x)=0 is x=(ln(3))/(ln(2))
Solve for x: log_{2}(4-2^x)=0
The solution is x=(ln(3))/(ln(2))