Logarithms question

  • Thread starter seiferseph
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  • #1
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Main Question or Discussion Point

if log2 = x and log3 = y, solve for log(base5)36 in terms of x and y.

how do i even get started on this? i'm really confused with logs.
 
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  • #2
Tide
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This seems like a rather roundabout way of doing things. In any case ...

I assume those logarithms are base 10 in which case, by definition of logarithm, we have

[tex]10^x = 2[/tex]
[tex]10^y = 3[/tex]

Let [itex]z = \log_5 10[/itex] which means [itex]5^z = 10[/itex]. But

[tex]5 = 2 + 3 = 10^x + 10^y[/tex]

so that

[tex]5^z = \left(10^x + 10^y\right)^z = 10[/tex]

IOW I'm not sure I see the point of the problem!

from which

[tex]z \log \left( 10^x + 10^z\right) = \log 10 = 1[/tex]

and finally

[tex]z = \frac {1}{\log \left( 10^x + 10^z \right)}[/tex]


Of course, a direct approach would have led to

[tex]\log_5 10 = \frac {\log 10}{\log_{10} 5} = \frac {1}{\log 5}[/tex]
 
  • #3
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thanks, but i'm not sure if that is correct (the teacher actually said it was quite simple). the last question was something like a^2 + b^2 for the answer. i'll post a little bit of what i got, i'm not sure if this is right
 

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  • #4
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Do you realize that your original question is about log_5(10), but the handwritten thing you posted now is about log_5(36)?
 
  • #5
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Muzza said:
Do you realize that your original question is about log_5(10), but the handwritten thing you posted now is about log_5(36)?
now i do :blushing: its supposed to be log_5(36)
 

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