How can the logarithm problem be solved?

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SUMMARY

The logarithm problem involving conjoined bases can be solved using the equation $$p^5=x$$ and $$p^2=y$$, leading to the conclusion that $$log_{xy}P = \frac{1}{7}$$. The derivation follows from the relationship $$(xy)^m = P$$ and simplifies to $$P^{7m}=P^1$$. The discussion highlights the importance of consistent notation when working with logarithmic identities, specifically $$\log_ab \ \log_ba=1$$ and $$\log_{xy}(p) = \frac{1}{2 + 5}$$.

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chwala
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Homework Statement
Given that, $$log_p X=5$$ and $$log_p Y=2$$.
Find $$log_{xy} P$$
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Interesting, i have not worked on logs with conjoined bases before, anyway my approach is as follows;

$$p^5=x$$ and $$p^2=y$$
Let $$log_{xy}P = m$$, →$$(xy)^m = P$$
$$(P^5⋅P^2)^m = P^1$$
$$P^{7m}=P^1$$
$$m=\frac {1}{7}$$

Any other way of looking at this is most welcome.
 
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As you find
\log_ab \ \log_ba=1
 
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\log_{xy}(p) = \frac{\log_p(p)}{\log_p(xy)} = \frac{1}{2 + 5}.
 
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You are, of course, assuming that x = X, y = Y and p = P. Or just being inconsistent in your notation.
 
I have been thinking about this question and my bad:mad:, i did not indicate the letters in the right manner. Find the question below as it appears on the textbook;

1640670828612.png


This implies that, my method was after all correct! Sorry for my lack of details here...
 
mjc123 said:
You are, of course, assuming that x = X, y = Y and p = P. Or just being inconsistent in your notation.
Kindly check my post ##5##...i did not capture the question correctly...My apologies mjc...
 

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