add logs to multiply, subtract to divide
logx (10*8/5) =4
logx (16) = 4
in general base^ logx = x
16 = x^4
x = 16^(1/4) = (2*2*2*2)^1/4 = 2
Log3 256-log3 27=x
same as last question, subtract is divide
log3 (256/27) = x
256/27 = 3^x
I bet you have a typo because 27 is 3^3 but 256 is nothing to the 3 so have to use calculator
9.48 = 3^x
log10 (9.48) = x log10 (3)
.9768 = x * .4771
x = .9768/.4771
base 10 I assume
Log 2+ 3log 5-log 25=x
log 2 + log 125 - log 25 = x
log (250/25) = x
log 10 = 1 = x
ok so by your previous help i tried to solve this problem
Log32 56-log32 7=x (log base 32)
what i did was 57/7=8
so in log form i got log 32 8=x ( log base 32)
then changed it to exponential form to 32^x=8
yes i do have a typo
Log32 56-log32 7=x
and my work is above
log 32 (8) = x
8 = 32^x
2^3 = 2^5x
x = 3/5 = 6/10 = 0.6 yes
I suspected you did not need a calculator :)
It helps if things like 2^3 = 8 and 2^5 = 32 are part of your tool box.
also very common
3^3 = 27
5^3 = 125
sorry to bug you but i think this may be the last problem
2log4 x=3 (log base 4)
i got log16^x=3
log4 (x^2) = 3
x^2 = 4^3 = 64
x = 8
log 32 x = 6/10
log 625 3/4 = x
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