How to calculate this log

Question

I'm supposed to calculate $$16^{\log_{0.5}{2.5}}$$

now the problem is it is not .25 rather 2.5. How can I solve this?

Answer 1

$16^{\log_{0.5}2.5}=16^{\ln2.5/\ln0.5}=16^{-\ln2.5/\ln2}=e^{-4\ln2.5}=2.5^{-4}$

Answer 2

First note $$\Large \log_{0.5} 2.5 = \log_{1/2}2.5=-\log_{2}2.5$$

We used $$\Large \log_{a^m}b=(1/m)\log_a b$$

So $$\Large 16^{-\log_{2}2.5}=2^{-4\log_{2}2.5}=(2.5)^{-4}$$

Answer 3

Working to base $2$ we have

$$16^{-\log 2.5}=(2^{\log 2.5})^{-4}=2.5^{-4}.$$

Answer 4

Just express the base $16$ in terms of $\frac 12$ and use laws of exponents:

$$16^{\log_{\frac 12}\frac 52}= 2^{4\log_{\frac 12}\frac 52}= \left( \left(\frac 12\right)^{\log_{\frac 12}\frac 52}\right)^{-4} $$$$= \left(\frac 52\right)^{-4}= \frac{16}{625}$$