Solving Logs: 3^x^3 = 9^x and 2^x = 10

[a_vanderbrook]

I understand how to basic logs, but then something like this problem...

1) 3^x^3 = 9^x
... how would I solve for this?

Also, I am having trouble with this problem...

2) 2^x = 10
.. any hints on how to start it off?


Thanks so much!

 

[a_vanderbrook]

I understand how to do*

 

[jwpaine]

Hey, I can help you on the second one.

2^x = 10

log both sides, so you have: xlog(2) = log(10) (base 10)

divide both sides by log(2)

x = log(10)/log(2)

 

[a_vanderbrook]

okay thanks a lot; i figured you would do that but i wasn't positive.
thanks for the help

 

[galactus]

For #1, do the usual log thing.

\(\displaystyle \L\\3^{x^{3}}=9^{x}\)

\(\displaystyle \L\\x^{3}log(3)=xlog(9)\)

\(\displaystyle \L\\x^{2}=\frac{log(9)}{log(3)}=\frac{2log(3)}{log(3)}\)

See now?.

 

[a_vanderbrook]

yes, thanks so much

 

[skeeter]

\(\displaystyle \L 3^{x^3} = 9^x\)

\(\displaystyle \L 3^{x^3} = 3^{2x}\)

\(\displaystyle \L x^3 = 2x\)

\(\displaystyle \L x^3 - 2x = 0\)

\(\displaystyle \L x(x^2 - 2) = 0\)

\(\displaystyle \L x(x + \sqrt{2})(x - \sqrt{2}) = 0\)

\(\displaystyle \L x = 0\)

\(\displaystyle \L x = \pm \sqrt{2}\)