okay will do, I tried but still stuck in verifying the answer. Could you please help and check what I missed?Continue with the quadratic you have found?
If a>b>1, which is bigger: log a (base b) OR log b (base a)?
This will help you determine which solution for the quadratic is which. And then you can complete the problem.
No.is log b (base a) equals to roughly 17?
how do I get log a (base b) though?
So what is the difference you were asked to solve. BTW, this is math so please stop with this approximation stuff. You need to find the exact answer. So far you have done very nice work.okay will do, I tried but still stuck in verifying the answer. Could you please help and check what I missed?
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What exactly do you mean by that, please?The same quadratic equation applies to both x = log b (base a) and x=log b (base a).
Because you seem adverse to using proper grouping symbols, I can only guess as to the correct reading the problem.Hello guys, I have problems and stuck with the question in the box. Would you please give directions to how I can solve the question? I have tried solving it below :
thank you
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I tried this but still trying to work it out, thanks for hints!I used the facts that 1/loga(b) = logb(a) along with logb(ab) = 1 + logb(a)
thanks for the walk through, i'll give it a try.Because you seem adverse to using proper grouping symbols, I can only guess as to the correct reading the problem.
Is it? Given: [imath]a>b>1~\&~\dfrac{1}{\log_a(b)}+\dfrac{1}{\log_b(a)}=\sqrt{293}[/imath], then find [imath]\dfrac{1}{\log_{ab}(b)}-\dfrac{1}{\log_{ab}(a)}=~?[/imath]
If that reading is correct then apply the change of base theorm: [imath]\log_j(k)=\dfrac{\log(k)}{\log(j)}[/imath]
So, then find [imath]\dfrac{1}{\log_{ab}(b)}-\dfrac{1}{\log_{ab}(a)}=\dfrac{\log(a)+\log(b)}{\log(b)}-\dfrac{\log(a)+\log(b)}{\log(a)}=~?[/imath]
the one on the right is bigger , but I am still confused how this will help meContinue with the quadratic you have found?
If a>b>1, which is bigger: log a (base b) OR log b (base a)?
This will help you determine which solution for the quadratic is which. And then you can complete the problem.
\(\displaystyle log_b{a}+\frac{1}{log_b{a}}=\frac{1}{log_a{b}}+log_a{b}\)What exactly do you mean by that, please?