Question

Find the exact value of each logarithm.
log1/2 16
can someone show me the steps

Answer 1

\[\left(\frac{1}{2}\right)^x=16\] solve for \(x\)

Answer 2

\[\log_\frac{1}{2} 16 \]\[=\frac{\log 16}{\log\frac{1}{2}}\]\[\log 16 = \log (2^4)\]\[\log \frac{1}{2} = \log (2^{-1})\]
Using \[\log m^n = n\log m\] You can express log16 and log(1/2) in terms of log2. Then, it should not be difficult.

Answer 3

\[\log 1^{-x}/(1/2)=16\]

Answer 4

\[\log 2^{-x}=16\]\[2^{-x}=16\]\[-x=4\]
x=-4

For some reason, I feel I have violated some mathematical rule in this approach. Please tell me if I erred.