I need help with solving logarithmic equations that have fractions, such as:

log8 n=4/3

*March 11, 2009, 1:01 pm by Guest*

log8 n=4/3

I need help with my intermediate algebra problems

*March 16, 2009, 2:29 pm by Guest*

I can not know exactly what do you want to solve. Did you mean to solve for "n" ? If so I can show you that :

Log (8n) = 4/3

(8n) = 10 ^ (4/3)

n = [10 ^ (4/3) ] / 8

This is the answer, you can use calculator to find the shorter answer. Good luck!

*March 29, 2009, 7:20 pm by Honest_denver09*

Log (8n) = 4/3

(8n) = 10 ^ (4/3)

n = [10 ^ (4/3) ] / 8

This is the answer, you can use calculator to find the shorter answer. Good luck!

If x = b^{y}, then y = log_{b} (x).

So in your problem log_{8}n = 4/3, x is n, b = 8 and y = 4/3.

So n = 8^{(4/3)}

n = 2^{(3*4/3)}

n = 2^{4} = 16

*June 19, 2009, 12:58 am by Guest*

So in your problem log

So n = 8

n = 2

n = 2

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