Question

Logarithm Question....


If y = \(a^{\left(\cfrac{1}{1-\log_a x}\right)} \) and z = \(a^{\left(\cfrac{1}{1-\log_a y}\right)}\)
then x = ?

Answer 1

If y = \(a^{\left(\cfrac{1}{1-\log_a x}\right)} \) and z = \(a^{\left(\cfrac{1}{1-\log_a y}\right)}\)

Answer 2

@Kainui and @iambatman

Answer 3

This is what I have done yet :

In order to get rid of the powers, I took log both sides :

\(\log y = \cfrac{1}{1-\log_a x} \log a\)

and

\(\log z = \cfrac{1}{1-\log_a y} \log a\)

Answer 4

Okay, let us try to simplify it more.

\(\log y = \left(\cfrac{1}{1-\log_a x} \right)\left(\cfrac{1 }{\log x}\right)\)

\(\log x \times \left(1-\log_a x\right) = \cfrac{1}{\log y}\)

\(\log x \times \left(1-\cfrac{\log a}{\log x }\right) = \cfrac{1}{\log y}\)

Answer 5

Seems like, I have made it more complicated.

Answer 6

This is what I have got after a lot of work :

\(\log_a \left(\cfrac{a}{x}\right) = \log_a \left(\cfrac{a}{y}\right) \times \left(\cfrac{\log y}{\log z}\right)\)

Answer 7

and making the bases same for everything :

\(\log_a \left(\cfrac{a}{x}\right) = \log_a \left(\cfrac{a}{y}\right) \times \left(\cfrac{\log_a z}{\log_a y}\right) \)

Answer 8

\(\cfrac{a}{x} = \left(\cfrac{a}{y}\right)^{\left(\cfrac{\log_a z}{\log_a y}\right)}\)

Answer 9

\(\log_a(z)=\frac{1}{1-\log_a(y)}\\1-\frac{1}{\log_a(z)}=\log_a(y)\\y=a^{\frac{\log_a(z-1)}{\log_a(z)}}\)
now

\(y=a^{\frac{1}{1-\log_a(x)}}\implies a^{\frac{\log_a(z-1)}{\log_a(z)}}=a^{\frac{1}{1-\log_a(x)}}\implies \frac{\log_a(z-1)}{\log_a(z)}=\frac{1}{1-\log_a(x)}\implies \\x=a^{\frac{\log_a(z)-\log_a(z-1)}{\log_a{z-1}}}\)

Answer 10

I think...

Answer 11

What you have written is though, not visible ot me, but yes, I'm getting what you did.

Answer 12

brb

Answer 13

I'm getting after a lot of disgusting looking algebra that if we use the same form of function,

x(z) has the same form as y(x) and z(y) shown above.

\[\Large x=a^{\frac{1}{1-\log_az}}\]

Answer 14

Well, may be these options help us :

\(\boxed{\\
\text{(A)} \quad a^{\cfrac{1}{1+\log_a z}} \\
\text{(B)} \quad a^{\cfrac{1}{2+ \log_a z}} \\
\text{(C)} \quad a^{\cfrac{1}{1-\log_a z}} \\
\text{(D)} \quad \text{None of these} \\ }\)

Answer 15

@Kainui - Yeah your answer seems to be right, as it is in the options.

Answer 16

I'm getting c, and here's my algebra for checking give me a moment...

Answer 17

I solved in general, \[\Large h=a^{\frac{1}{1-\log k}} \]\[\Large \log h=\frac{1}{1-\log k}\]\[\Large 1-\log k=\frac{1}{\log h}\]\[\Large 1-\frac{1}{\log h}= \log k\]\[\Large k=a^{ 1-\frac{1}{\log h}}\]

Now since h(k) is the same form as both y(x) and z(y) we can plug in each equation to itself... One sec...

Answer 18

We now have found k(h) so we really have x(y) and y(z) now so we simply find x(y(z)) since that's what all our answer choices are.

Answer 19

\[\Huge x(y(z))=a^{ 1-\frac{1}{\log (a^{ 1-\frac{1}{\log h}})}}\]The log and a part in the exponent cancel out.

\[\Huge x=a^{ 1-\frac{1}{ 1-\frac{1}{\log h}}}\] From here we simplify the fractions

\[\Huge x=a^{ 1-\frac{1}{ \frac{\log h -1}{\log h}}}\]\[\Huge x=a^{ 1-\frac{\log h}{ \log h -1}}\]\[\Large x=a^{ \frac{\log h -1 -\log h}{ \log h -1}}\]\[\Large x=a^{ \frac{-1}{ \log h -1}}=a^{ \frac{1}{ 1-\log h }}\]

But I can't help but feel like there's a better way of doing this one.

Also I just realized all those h's should be z's. Whatever.

Answer 20

Well, yeah, I think this is one of the ways of solving this quesiton,,,, but there must be an easier way to get to the answer.

Answer 21

Obviously there's a symmetry going on here, but it beats me how it works. Really cool problem.

Answer 22

This is what we can do :

\((1-\log_a x)(\log y) = (1-\log_a y)(\log z)\)

Answer 23

I think I got it.

Answer 24

What is it? I think there's a really nice solution to this that's really fast.

Answer 25

This might be better:

h(k) has the form of y(x) and z(y)
k(h) has the form x(y) and y(z). So we can set y(x)=y(z) if we just play around with the general equation.

\[\Large h=a^{1-\frac{1}{\log k}}\]
Ok so skipping the little algebra I get to these two equations from this one. \[\Large \log h=1-\frac{1}{\log k}\]\[\Large \log k=\frac{1}{1-\log h}\]
Above when I said y(x)=y(z) this also means log[y(x)]=log[y(z)] So let's do that.
\[1-\frac{1}{\log x}=\frac{1}{1-\log z}\]
Solve for x. I don't know if this is really that much better hmm... I feel like there's a really satisfying way to recognize that y(x), z(y) and x(z) are all identical forms without having to do anything at all.

Answer 26

Sorry, my net stopped working.

Answer 27

Yeah, I agree Kainui that there should be much satisfying way to do this.