Question 973713: Did I evaluate these correctly?
a) log5(6) + log3(7) = log5(42)
b) b) 3 log2(16)
= log2(16)3
= 4^3
= 64
c) Z = 10 logI(1) - 10 logI(0)
= 10 logI(1/0) = 10 logI(0)
d) 124 = 3(4)^2x-1
log 124 = log 12^(2x-1)
log 124 / log12 = 2x - 1
10.3... = 2x - 1
10.3... +1 = 2x -1 +1
11.3... = 2x
(11.3/2) = (2x/2)
5.66... = t
e) 12^(t-4) = 17^(-3t)
log12^(t-4) = log17^(-3t)
t-4 log12 = -3t log17
-4t log12 = -3t log17
(-4t) + 4t = (-3t) log 17 + 4t
log12 = t log17
log12/log 17 = t
0.77... = t Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Did I evaluate these correctly?
a) log5(6) + log3(7) = log5(42)
No, the logs have different bases unless it's a typo.
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b) b) 3 log2(16)
= log2(16)3
= 4^3
= 64
-------------
16 = 2^4
3*4 = 12
----------------------------------
c) Z = 10 logI(1) - 10 logI(0)
= 10 logI(1/0) = 10 logI(0)
Not clear what you mean by logI(...)
You can't use log(0) in any case.
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d) 124 = 3(4)^2x-1
Assuming the exponent of 4 is 2x-1:
log(124) = log(3) + (2x-1)*log(4)
log(124) - log(3) = (2x-1)*log(4)
log(124/3) = (2x-1)*log(4)
2x-1 = log(124/3)/log(4)
2x = log(124/3)/log(4) + 1
x = log(124/3)/(2log(4)) + 1/2
x = log(124/3)/log(16) + 1/2
============================================
log 124 = log 12^(2x-1)
log 124 / log12 = 2x - 1
10.3... = 2x - 1
10.3... +1 = 2x -1 +1
11.3... = 2x
(11.3/2) = (2x/2)
5.66... = t
e) 12^(t-4) = 17^(-3t)
-----------------
ooooh, way off.
(t-4)*log(12) = -3t*log(17)
tlog(12) - 4log(12) = -3tlog(17)
tlog(12) + 3tlog(17) = 4log(12)
t(log(12) + 3log(17)) = log(20736)
t = log(20736)/log(12*17^3)
t =~ 0.90487
==============================================
log12^(t-4) = log17^(-3t)
t-4 log12 = -3t log17
-4t log12 = -3t log17
(-4t) + 4t = (-3t) log 17 + 4t
log12 = t log17
log12/log 17 = t
0.77... = t
===========================================
Whether you get it right or not, I always give priority to someone who tries.
