2x = 25
x=5
82 = 64
log864 = 2
log24 = 2
22 = 4
log2x = log210
x = 10
Given log3=5 and log2=3, what is log15?
log15= 5+3 = 8
52x = 520
x = 10
172 = 289
log17289 = 2
log864 = 2
82 = 64
log5x = log2x-9
x = -3
Given log2=3 and log5=2, what is the value of log5/2?
log5/2= 3-2 = 1
33x = 92x-6
2561/2 = 16
log25616 = 1/2
log5(1/25) = -2
5-2 = 1/25
log(-5x-6) = log(x2)
x = -2
x = -3
Given log2=3 and log3=4, what is the value of log12?
log12 = (log2)(log2)(log3) = 3+3+4 = 10
4x = 82x+5
x = 15/4 OR 3.75
(1/6)3 = (1/216)
log(1/6)(1/216) = 3
log(1/8)64 = -2
(1/8)-2 = 64
log2x-log210 = 2
x = 40
Given log3=5, log2=3 and log5=7, what is the value of log(12/5)?
log(12/5) = (log2)(log2)(log3)/(log5) = 5+3+3-7 = 4
16x-5 = (1/32)x-6
x = -5
64-1/2 = 1/8
log64(1/8) = -1/2
log100 = 2
102 = 100
2log54+log5x = 3
125/16 OR 7.8125
Given log2=5 and log3=2, is it possible to find the value of log42?
No