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I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple.

- Thread starter cscott
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- #1

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I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple.

- #2

Curious3141

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What you have isn't that much of a simplification, it would be better to break everything on the LHS to powers of 2 and get smaller numbers. Nevertheless, you can solve the equation immediately from what you have.cscott said:

I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple.

What is the definition of a logarithm ? What does it mean when you say [tex]log_a b = c[/tex]. Come up with a simple equation to relate a, b and c using the

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x^(LOGx32)=x^5 the x and the LOGx cancel out the we have

=>32=x^5

therefore x=2

- #4

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[tex]20 000 = 10 000(x)^9[/tex]

[tex]2 = x^9[/tex]

[tex]log 2 = log x^9[/tex]

[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...

- #5

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Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?steven187 said:

x^(LOGx32)=x^5 the x and the LOGx cancel out the we have

=>32=x^5

therefore x=2

- #6

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Actually, once you've got [tex]2 = x^9[/tex], all you've got to do is [tex]\sqrt[9]{2} = x = 1.080059739[/tex].cscott said:

[tex]20 000 = 10 000(x)^9[/tex]

[tex]2 = x^9[/tex]

[tex]log 2 = log x^9[/tex]

[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...

I don't think it's really trial and error; had it been 31 or 33, the answer wouldn't have been an integer. "Luckily", [tex]\sqrt[5]{32} = 2[/tex].cscott said:Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?

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James R

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[tex]\log_a b[/tex]

is the power which you must raise a to in order to get b. So

[tex]\log_2 32 = 5[/tex]

Equivalently, if

[tex]\log_a b = x[/tex]

then

[tex]b = a^x[/tex]

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