Posted by **Jus** on Saturday, May 23, 2009 at 7:23pm.

Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0

- Math -
**bobpursley**, Saturday, May 23, 2009 at 8:17pm
take the log of each side

loga=(loga/logb)logb

now reduce.

- Math -
**Jus**, Saturday, May 23, 2009 at 8:43pm
I have:

loga=(loga/logb)logb

(loga/logb )(1/loga)=logb

(1/logb)=logb

Ok now what?

- Math -
**bobpursley**, Saturday, May 23, 2009 at 10:13pm
you erred.

loga=(loga/logb)logb

the logb on the right side divide out (one on numerator, one in denominator)

loga=loga

divide both sides by loga.

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