Log problems

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nanase

Junior Member
Joined
Aug 8, 2019
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232
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
Log question.jpeg
 
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.
 
Harry_the_cat said:
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.
Click to expand...
okay will do, I tried but still stuck in verifying the answer. Could you please help and check what I missed?
log ans.jpeg
 
is log b (base a) equals to roughly 17?
how do I get log a (base b) though?
 
nanaseailie said:
is log b (base a) equals to roughly 17?
how do I get log a (base b) though?
Click to expand...
No.
The same quadratic equation applies to both x = log b (base a) and x=log b (base a). Can you confirm that?
So your solutions for the quad eqtn : one is log b (base a) and the other is log a (base b).
Given that a>b>1, which is which?
And then sub into the final expression.
 
nanaseailie said:
okay will do, I tried but still stuck in verifying the answer. Could you please help and check what I missed?
View attachment 34457
Click to expand...
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.
 
nanaseailie said:
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
View attachment 34456
Click to expand...
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]
 
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I used the facts that 1/loga(b) = logb(a) along with logb(ab) = 1 + logb(a)
 
pka said:
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]
Click to expand...
thanks for the walk through, i'll give it a try.
 
Harry_the_cat said:
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.
Click to expand...
the one on the right is bigger , but I am still confused how this will help me
 
logab + logba = sqrt(293)
logab + 1/logab = sqrt(293)
Let x=logab
Now you have x + 1/x = sqrt(293) which can be expressed as a quadratic in x and therefore you can solve for x.
That is, we now have a value for logab


logabb - logaba = 1/logb(ab) - 1/loga(ab) = 1/[1+logba] - 1/[1+logab] = 1/[1+1/logab] - 1/[1+logab]

Now you finish up.
 
Steven G said:
What exactly do you mean by that, please?
Click to expand...
\(\displaystyle log_b{a}+\frac{1}{log_b{a}}=\frac{1}{log_a{b}}+log_a{b}\)

So the quadratic equation found in post #1 applies to both \(\displaystyle x = log_a{b}\) and \(\displaystyle x=log_b{a}\).

So, given that \(\displaystyle a>b>1\), it follow that \(\displaystyle log_a{b} < log_b{a}\) , so you can work out which root is which. This is important in working out the correct sign for your answer.
 
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