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Posted by **johnica martin** on Sunday, April 4, 2010 at 6:50pm.

Logx10 + logx8-logx5=4

Log3256-log327=x

Log 2+ 3log 5-log 25=x

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**Damon**, Sunday, April 4, 2010 at 6:53pmadd 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

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**Damon**, Sunday, April 4, 2010 at 7:00pmLog3 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

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**Damon**, Sunday, April 4, 2010 at 7:02pmbase 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

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**johnica martin**, Sunday, April 4, 2010 at 7:14pmok 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?

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**johnica martin**, Sunday, April 4, 2010 at 7:15pmyes i do have a typo

Log32 56-log32 7=x

and my work is above

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**Damon**, Sunday, April 4, 2010 at 7:22pmlog 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 :)

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**Damon**, Sunday, April 4, 2010 at 7:27pmIt helps if things like 2^3 = 8 and 2^5 = 32 are part of your tool box.

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**Damon**, Sunday, April 4, 2010 at 7:29pmalso very common

3^3 = 27

5^3 = 125

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**johnica martin**, Sunday, April 4, 2010 at 7:55pmsorry 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?

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**Damon**, Sunday, April 4, 2010 at 9:01pmlog4 (x^2) = 3

x^2 = 4^3 = 64

x = 8

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**Blake**, Wednesday, June 2, 2010 at 5:28pmlog 32 x = 6/10

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**Blake**, Wednesday, June 2, 2010 at 5:29pmlog 625 3/4 = x