Question

\[(\log_{x} 3x)(\log_{3}x ^{2})=(\sin1/\cos1)(\sin2/\cos2)(\sin3/\cos3)...(\sin89/\cos89)\]

Answer 1

First remember that sin(x) = cos(90-x). This should enable you to radically simplify the RHS of this equation

Answer 2

so should it be (cos(90-x)/cos1)? I'm not sure exactly what to do...

Answer 3

For example, using that identity, sin 1 = cos 89. Hence that pair of terms cancel. Now what about all of the other terms on the RHS (right hand side)?

Answer 4

I got the first part being (log3x/logx)\[(2\log_{3}x/\log3)\]

Answer 5

therefore the (sin1/cos1)=(cos89/cos1)=cos89? am i understanding that correctly?

Answer 6

sin 1 = cos 89. Hence sin 1 / cos 89 = 1. Every other pair of terms cancel out as well:
i.e.,
sin 2 / cos 88 = 1
sin 3 / cos 87 = 1
...

Answer 7

so would the next one be (cos88)?

Answer 8

next what? Do you understand what I just wrote above?

Answer 9

not completely. aren't you replacing sin1=cos89 therefore wouldn't it look something like \[\cos 89/\cos 1 => \cos 89\]

Answer 10

(sin1/cos1)(sin2/cos2)(sin3/cos3)...(sin89/cos89)
= (sin1/cos89)(sin2/cos88)....(sin88/cos2)(sin89/cos1)
by just rearranging terms.

Now each of these fractions is equal to exactly 1.

Answer 11

what about the first part of the problem with the logs? can you help me understand what to do next?

Answer 12

I would change base to make the logs all be the same. Here I am changing them all to base \( e\) and hence writing \( \ln = \log_e \)\[ (\log_x 3x)(\log_3 x^2) = \frac{ \ln 3x}{\ln x} \frac{\ln x^2}{\ln 3}\]

Now simplify this expression.

Answer 13

i don't know how. :(

Answer 14

this was the farthest I've solved it: (log3x/logx)(2log3x/log3)

Answer 15

Here's one more hint then: \( \ln x^2 = 2 \ln x \). Now notice you can do some cancelation.

Answer 16

so it comes out to 2x