The following equation needs to be done without the use of a calculator and for the life of me I can't figure out how. If anyone has an idea of how to do it, please let me know. Log bases are in square brackets here.
4^(0.5log[4](9) - 0.25log[2](25))
The answer is apparently 3/5 but I can't figure out how to do it without the use of a calculator. Thanks.
+118677 I really appreciate how well you have presented your problem. 
People often leave brackets out with questions like these and they become very ambiguous.
I won't claim that i have done it the easiest way. This was a difficult one. But the answer is correct.
4^(0.5log[4](9) - 0.25log[2](25))
$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\
=4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\
=4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\
=4^{log_{4}\;3-log_2\;5^{0.5}}\\\\
=4^{log_{4}\;3-0.5log_2\;5}\\\\$$
Now, I can't do this unless I can get the bases the same.
$$\begin{array}{rll}
let\;\; y&=&log_2 5\\\\
5&=&2^y\\\\
5&=&4^{0.5y}\\\\
log_4 5&=&log_4 4^{0.5y}\\\\
log_4 5&=&0.5ylog_4 4\\\\
log_4 5&=&0.5y\\\\
y&=&2log_4 5\\\\
log_2 5&=&2log_4 5\\\\
\end{array}$$
-------------------------------
so
$$=4^{log_{4}\;3-0.5log_2\;5}\\\\
=4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\
=4^{log_{4}\;3-log_4\;5}\\\\
=4^{log_{4}\;(3/5)}\\\\
=\frac{3}{5}$$
calculator check - using the web2 site calculator.
$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$
The calc has a little rounding error - the answers are the same.
+118677 I really appreciate how well you have presented your problem.
People often leave brackets out with questions like these and they become very ambiguous.
I won't claim that i have done it the easiest way. This was a difficult one. But the answer is correct.
4^(0.5log[4](9) - 0.25log[2](25))
$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\
=4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\
=4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\
=4^{log_{4}\;3-log_2\;5^{0.5}}\\\\
=4^{log_{4}\;3-0.5log_2\;5}\\\\$$
Now, I can't do this unless I can get the bases the same.
$$\begin{array}{rll}
let\;\; y&=&log_2 5\\\\
5&=&2^y\\\\
5&=&4^{0.5y}\\\\
log_4 5&=&log_4 4^{0.5y}\\\\
log_4 5&=&0.5ylog_4 4\\\\
log_4 5&=&0.5y\\\\
y&=&2log_4 5\\\\
log_2 5&=&2log_4 5\\\\
\end{array}$$
-------------------------------
so
$$=4^{log_{4}\;3-0.5log_2\;5}\\\\
=4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\
=4^{log_{4}\;3-log_4\;5}\\\\
=4^{log_{4}\;(3/5)}\\\\
=\frac{3}{5}$$
calculator check - using the web2 site calculator.
$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$
The calc has a little rounding error - the answers are the same.
+1313 I think there is easier way. I'm going to work on this for my own revision. Those logs are what I always forget how to and what to and where to, but don't wait for an answer. Good luck.
@Melody
Jesus, that was amazing. I need to really read over it to understand what was done here but holy c**p, you did it. It's possible. Thanks so, so much :)
+118677 You are very welcome 
I really appreciate your enthusiasm but even so a little less swearing would be good.
I am sure that you do not want to offend anyone.
Melody - noted, and thanks again
Stu - if you have an easier way I'd really like to see it so I'll stay posted on this page if you ever get around to it.
+33661 Here's a slightly different approach, though it ultimately amounts to the same as Melody's:
+129852 Here's my (belated) take on this one:
We can write:
4^(0.5log4(9) - 0.25log2(25)) ...as.....
[4^log4(3)] / [ 4^ log2(5)(.5)]
4log4(3)/ [4(log2(5)/2)] ...... the numerator simplifies to 3
Note that log2(5) is just a number.....call it "a' ....so we have
4(a/2) = [4^(1/2)]^a = 2(a) =
2log2 5 = 5
So our answer is just ..... 3/5
Sorry I haven’t gotten back sooner. I was so amazed by Melody’s answer I didn’t check back to see if anyone else posted. Two more great answers.
Melody’s was like a dissection to find the answer. Alan’s was like a resection.
CPhill’s answer was like he chopped it up with a machete and found the answer hidden inside. Jesus Christ, CPhill, you really are fucking amazing! You are like the guy I watched whacking the s**t out of a coconut with an axe and rock, then after a few minutes, my grandmother’s face appeared.
Grandma didn’t think it looked like her but everyone else did. After she bitched for awhile, my brother says, well Grandma, if you’d go to a plastic surgeon he might make you look as good as the coconut. My other brother says, s***w that, send her to the guy with the axe and rock, he does great work and he’s cheap. Everyone thinks this is hilarious, except Grandma, of course. She got really pissed about it.
If I, or any of my friends have another problem like this, you can bet your sweet a*s I’ll send them here.
Thank you all, so very much.
Charlotte
