SOLUTION: evaluate
e^2ln3
(2^3*2^-5)/2^-7
log4 64
log4 32
log2 4th root(1/8)
{{{(3^-2+3^0)^-2}}}
log 25 +log 4
Algebra.Com
Question 66208: evaluate
e^2ln3
(2^3*2^-5)/2^-7
log4 64
log4 32
log2 4th root(1/8)
log 25 +log 4
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
evaluate
e^2ln3 = [e^(ln3)]^2 =[3]^2 = 9
----------------------
(2^3*2^-5)/2^-7 = 2^-2 / 2^-7 = 2^5 = 32
------------------
log4 64 = log4 (4^3) = 3
---------------
log4 32 = log2 32 / log2 4 = 5/2
-------------
log2 4th root(1/8) = (1/4)log2 [2^-3] = (1/4)(-3) = -3/4
---------------
log 25 +log 4 = log (25*4)= log 100 = 2
-----------------
Cheers,
Stan H.
RELATED QUESTIONS
1/2 log4 64 - 4 + 3 log4 2... (answered by ikleyn)
log4^2 + log4^8 =... (answered by ankor@dixie-net.com)
express log4^2+log4^32 in single log and... (answered by Fombitz)
Given log 4^(y-1)+ log4^(8/2) = a+log2^(y+1)-log2^3=a-1. Show that... (answered by greenestamps)
solve for log(3x-1) + log2= log4 +... (answered by lwsshak3)
Without using tables, simplify
[(log3/2)+(log4/5)-log30] ÷... (answered by josmiceli)
log4(2)−log4(x)=log4(3) (answered by jim_thompson5910)
how to solve logx 49/64= -2
solve for x:
logx 2=1/3
simplify log2 2^2=log5... (answered by lwsshak3)
2 log4 (x) +3 log 8... (answered by jsmallt9)