# math

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Logx10 + logx8-logx5=4

Log3256-log327=x

Log 2+ 3log 5-log 25=x

• math -

add logs to multiply, subtract to divide
logx (10*8/5) =4

logx (16) = 4

in general base^ logx = x
so
16 = x^4
x = 16^(1/4) = (2*2*2*2)^1/4 = 2

• math -

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
anyway
9.48 = 3^x
log10 (9.48) = x log10 (3)
.9768 = x * .4771
x = .9768/.4771

• math -

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

• math -

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
x= 0.6
correct?

• math -

yes i do have a typo
Log32 56-log32 7=x
and my work is above

• math -

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 :)

• math -

It helps if things like 2^3 = 8 and 2^5 = 32 are part of your tool box.

• math -

also very common
3^3 = 27
5^3 = 125

• math -

sorry to bug you but i think this may be the last problem
2log4 x=3 (log base 4)
i got log16^x=3
16^3=x
x=4096
right?

• math -

log4 (x^2) = 3
x^2 = 4^3 = 64
x = 8

• math -

log 32 x = 6/10

• math -

log 625 3/4 = x

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