1. ## log problem

Given that $\log_4T=2+\log_2v,$express t in term of V.

2. Originally Posted by mastermin346
Given that $\log_4T=2+\log_2v,$express t in term of V.

hmmmm.....

$\log_4T=2+\log_2v,$

$T=4^{2+\log_2v}$

3. Hello, pickslides!

I just noticed that your answer can be simplified beyond all recognition . . .

$T\:=\:4^{2+\log_2v}$

$T \;=\;4^2\cdot 4^{\log_2v} \;=\;16\cdot(2^2)^{\log_2v} \;=\;16\cdot2^{2\log_2v} \;=\;16\cdot2^{\log_2(v^2)} \;=\;16v^2$

Yeah, you're right . . . I've got far too much time on my hands.
.

4. Hello,

Here is another way :

We know that $\log_a b=\frac{\ln b}{\ln a}$
From here, it's easy to see that $\log_2 v=2 \log_4(v^2)$

And we know that $2=\log_4 4^2=\log_4 16$, so that $2+\log_2 v=\log_4 (16v^2)$

---> $\log_4 T=\log_4 (16v^2)\Rightarrow T=16v^2$