What is the answer to log_{2}(x+1)=log_{3}(34.67) ?

The answer to log_{2}(x+1)=log_{3}(34.67) is x=2^{log_{3}(34.67)}-1

Solve for x: log_{2}(x+1)=log_{3}(34.67)

The solution is x=2^{log_{3}(34.67)}-1

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lo g_{2} (x + 1) = lo g_{3} (34.67)

x = 2^{l o g_{3} (34.67)} - 1

Decimal Notation

x = 8.36704 \dots

Solution steps

lo g_{2} (x + 1) = lo g_{3} (34.67)

Apply log rules

x + 1 = 2^{l o g_{3} (34.67)}

Solve

x + 1 = 2^{l o g_{3} (34.67)} : x = 2^{l o g_{3} (34.67)} - 1

x = 2^{l o g_{3} (34.67)} - 1

lo g_{3} (2 x + 1) - lo g_{3} (x - 2) = 1

x (2 ln (x) - 1) = 0

lo g_{4} (x - 3) + lo g_{4} (x + 60) = 3

lo g_{x} (x) = 0.4

ln (2) + ln (x - 3) = 4

What is the answer to log_{2}(x+1)=log_{3}(34.67) ?
The answer to log_{2}(x+1)=log_{3}(34.67) is x=2^{log_{3}(34.67)}-1
Solve for x: log_{2}(x+1)=log_{3}(34.67)
The solution is x=2^{log_{3}(34.67)}-1