Log Based Question

[mthcs]

I had a question regarding this.... 9+7lnx=5

If i move the 9 over I will get -4 so I am not sure what to do. Thanks in advance.

And another question.. for the equation e^9-7x=6469 how do I solve this one??

Thank you

 

[mthcs]

Ok i just had one other question... the question is 8^x=1/(the 9th root of 2) AKA 8^x=(1)/(2^(1/9))

I don't get how to even start this one.

Thanks!

 

[pka]

Ok i just had one other question... the question is 8^x=1/(the 9th root of 2) AKA 8^x=(1)/(2^(1/9))

Lets play the "back-of-the-book" game.
Look it up and find \(\displaystyle x=\dfrac{-1}{27}\). Now you need to find out HOW & WHY?
Hint: \(\displaystyle \ln(8)=3\ln(2)~\&~9\cdot 3=27\)

 

[mthcs]

Thank you. I get that it would turn into 3x=log(base 2)(1/2^(1/9) but I am not sure how to get a direct x value from there.

 

[pka]

Thank you. I get that it would turn into 3x=log(base 2)(1/2^(1/9) but I am not sure how to get a direct x value from there.

Since you have already been given a detailed solution, I will show you an easy way.
Know that \(\displaystyle a^x=a^y,~a>0\) if and only if \(\displaystyle x=y\).
Now your question reduces to \(\displaystyle 2^{3x}=2^{(-1/9)}\) thus \(\displaystyle 3x=-1/9\;.\)