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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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Do you mean $13^{4x} - 5 = 6$ or $13^{4x - 5} = 6$?

Either way, actually, it's just a logarithm. To get the 4x "down" you need to take the logarithm of both sides.

Remember that $\log_3 (3^x) = x$ and so on.

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

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

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No. You'll also need the change of base formula if your calculator doesn't let you choose your own base.

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I don't quite understand...

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

$\log_{13}(13^{4x-5}) = 4x -5$

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

$\log_n(a) = \frac{\log_{10}(a)}{\log_{10}(n)}$

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