log(y) / log(x) + log(z) / log(y) + log(t)/ log(z) =8.x2.t


Ican't find x Can you help me , please ? Thanks

What do YOU think ?
While we're happy to HELP we wont do all the work for you.
Show us what you have done and where you are having problems..

What do YOU think ?
While we're happy to HELP we wont do all the work for you.
Show us what you have done and where you are having problems..

log t /log x =8x^2t I have already done but I can't understand the rest Thank you

Please confirm that what you meant is this:

\frac {\log y}{\log x}\ +\ \frac {\log z }{\log y} \ +\ \frac {\log t}{\log z} = 8 x^2 t

Is this exactly how the problem was presented to you? Because this is an unusual problem, and in fact I don't believe it has a unique solution, so before going too far down the wrong path I just want to make sure we are on the same page.

Just as a guess - I wonder if the original problem was actually this:

\log (\frac y x) + \log (\frac z y )+ \log (\frac t z ) = \log (8 x^2 t)

Is my guess correct?

[QUOTE=ebaines;3761600]Please confirm that what you meant is this:

\frac {\log y}{\log x}\ +\ \frac {\log z }{\log y} \ +\ \frac {\log t}{\log z} = 8 x^2 t

Is this exactly how the problem was presented to you? Because this is an unusual problem, and in fact I don't believe it has a unique solution, so before going too far down the wrong path I just want to make sure we are on the same page.

Just as a guess - I wonder if the original problem was actually this:

\log (\frac y x) + \log (\frac z y )+ \log (\frac t z ) = \log (8 x^2 t)

Hi Ebaines,
Unfortunately my question is correct . The result should be 1, 2, 3, 4 or 5
Thanks.

Hi Ebaines,
Unfortunately my question is correct . The result should be 1, 2, 3, 4 or 5
Thanks

Unfortunately the correct answer is "none of the above." Of course x cannot = 1, because then the log(x) = 0 and the first term would have a zero in the denominator. Other than that - it can be shown that the value for 'x' is not defined unless the values for y, z, and t are set. So there is not a unique answer to the problem as written.

Thanks for your help...