How to evaluate this expression?
$$ \ 3^{\log_4 5} + 4^{\log_5 3} - 5^{\log_4 3} - 3^{\log_5 4} \ $$
Asked
Active
Viewed 108 times
1
-
2Hint: $;a^{log_b(c)} = c^{log_b(a)},$. – dxiv Jul 02 '17 at 06:00
-
1How? Is it any specific theorem? – Fghj Jul 02 '17 at 06:01
-
Take the $\log$ on both sides and use that $\log_u(v)=\log(v) / \log(u),$. – dxiv Jul 02 '17 at 06:02
-
1Where are the two sides? This is just an expression. Can I equate it to 0 ? – Fghj Jul 02 '17 at 06:04
-
I was referring to the equality in my previous comment. – dxiv Jul 02 '17 at 06:04
-
2$$ 3^{\log_4 5} = \left( 4^{\log_4 3} \right)^{\log_4 5} = 4^{(\log_4 3 \cdot \log_4 5)} = \left( 4^{\log_4 5} \right)^{\log_4 3} = 5^{\log_4 3}. $$ – Michael Hardy Jul 02 '17 at 06:06
1 Answers
3
Very often when you encounter an expression like $g(x)^{f(x)}$ you would want to to rewrite it as $e^{f(x)\ln(g(x))}$.
Also, remember that $\log_a(b) = \frac{\ln(b)}{\ln(a)}$.
With that in mind: $3^{\log_4(5)} = e ^ { \frac{\ln(5) }{ \ln(4)} \ln(3) } $ and same is for $5^{\log_4(3)} = e ^ { \frac{\ln(3) }{ \ln(4)} \ln(5) } $ so they cancel each other.
Same is for other two summands. So the whole expression evaluates just to zero.
D. Dmitriy
- 993
- 4
- 8