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quick exponential question

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how would I solve this?

 

13^4x-5 = 6

 

because I thought that 13 and 6 would have had to have a common power...

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woops i mean 13^(4x-5)=6

 

so would i go log(4x-5)13=log6?

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I don't quite understand...

You'd have to take the base 13 logarithm of both sides.

 

[math]\log_{13}(13^{4x-5}) = 4x -5[/math]

 

If you calculator can't do base 13 logs (to find [math]\log_{13}(6)[/math]), you can do this:

 

[math]\log_n(a) = \frac{\log_{10}(a)}{\log_{10}(n)}[/math]

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