[MOVED] proof for logs: if log_a(b) = p^3 and....

[Guest]

If log_a(b)= p^3 and log_b(a) = 4/ p^2, show that p = 1/4.

I need help with this proof please, thank you. I have trouble getting these.

 

[galactus]

Note the a and b are just reversed in the logs.


\(\displaystyle \L\\\frac{1}{p^{3}}=\frac{4}{p^{2}}\)

Solve for p.

 

[stapel]

Hint: Use the change-of-base formula:

. . . . .\(\displaystyle \L \log_a{(b)}\,=\,\frac{\log_b{(b)}}{\log_b{(a)}}\,=\,\frac{1}{\log_b{(a)}}\)

Then substitute to get the equation provided in the earlier reply.

Eliz.

 

[soroban]

Hello, bittersweet!

\(\displaystyle \text{Given: }\log_a(b)\,=\,p^3\text{ and }\log_b(a)\,=\,\frac{4}{p^2}\;\;\text{ Show that: }\,p\,=\,\frac{1}{4}\)

Galactus has the best approach.

Note that \(\displaystyle \L p\,\neq\,0\).


Theorem \(\displaystyle \L\;\log_a(b)\:=\:\frac{1}{\log_b(a)}\)

So we have: \(\displaystyle \L\,p^3\:=\:\frac{1}{(\frac{4}{p^2})}\;\;\Rightarrow\;\;p^3\:=\:\frac{p^2}{4}\;\;\Rightarrow\;\;4p^3\,-\,p^2\:=\:0\)

Then: \(\displaystyle \L\,p^2(4p\,-\,1)\:=\:0\;\;\Rightarrow\;\;\sout{p \,=\,0},\;\;p\,=\,\frac{1}{4}\)


Therefore: \(\displaystyle \L\,p\,=\,\frac{1}{4}\)