Logarithms

Jan 2010
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Not sure how to put this in here, so hopefully i can do my best
Please help with this

\(\displaystyle log3(logx(log416))=-1\)
The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right
 

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MHF Hall of Honor
Mar 2010
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Not sure how to put this in here, so hopefully i can do my best
Please help with this

\(\displaystyle log3(logx(log416))=-1\)
The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right
Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

\(\displaystyle log_3(log_x(log_416))=-1\)

\(\displaystyle log_x(log_416)=3^{-1}=\frac{1}{3}\)

\(\displaystyle log_416=x^{\frac{1}{3}}\)

\(\displaystyle 2=x^{\frac{1}{3}}\)

\(\displaystyle x=2^3 = 8\)
 
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Jan 2010
60
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Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

\(\displaystyle log_3(log_x(log_416))=-1\)

\(\displaystyle log_x(log_416)=3^{-1}=\frac{1}{3}\)

\(\displaystyle log_416=x^{\frac{1}{3}}\)

\(\displaystyle 2=x^{\frac{1}{3}}\)

\(\displaystyle x=2^3 = 8\)
wow, that was fast and easy, i couldn't remember how the bases work, but i do know, boot the log, thanks so much for your help.
 
Jan 2010
60
0
so for this one, i got an answer of x=5, is this correct

\(\displaystyle _62log_6x+log_6x=125\)
 

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MHF Hall of Honor
Mar 2010
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so for this one, i got an answer of x=5, is this correct

\(\displaystyle _62log_6x+log_6x=125\)
Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?
 
Jan 2010
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Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?
the 2log6x+log6x is all raised, not sure how else to explain it
 

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MHF Hall of Honor
Mar 2010
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the 2log6x+log6x is all raised, not sure how else to explain it
Is it like this?

\(\displaystyle \displaystyle{6^{2log_6x+log_6x}=125}\)

(I know it's kind of hard to read, but if it's right you should be able to tell.)
 
Jan 2010
60
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Is it like this?

\(\displaystyle \displaystyle{6^{2log_6x+log_6x}=125}\)

(I know it's kind of hard to read, but if it's right you should be able to tell.)
Thats exactly how it's written, good job
 

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MHF Hall of Honor
Mar 2010
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Chicago
Thats exactly how it's written, good job
\(\displaystyle \displaystyle{6^{2\log_6x+\log_6x}=125}\)

\(\displaystyle 2\log_6x+\log_6x=\log_6(125)\)

\(\displaystyle 3\log_6x=\log_6(125)\)

\(\displaystyle \log_6(x^3)=\log_6(125)\)

\(\displaystyle x^3=125\)

\(\displaystyle x=5\)

So, to answer your question, yes. :)
 
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Jan 2010
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\(\displaystyle \displaystyle{6^{2\log_6x+\log_6x}=125}\)

\(\displaystyle 2\log_6x+\log_6x=\log_6(125)\)

\(\displaystyle 3\log_6x=\log_6(125)\)

\(\displaystyle \log_6(x^3)=\log_6(125)\)

\(\displaystyle x^3=125\)

\(\displaystyle x=5\)

So, to answer your question, yes. :)
awesome, exactly what i got, just wanted to double check, some of these logs are so confusing, thanks alot