a fun(?) log Q: Given a,b>1, prove 1/log_b(a) + ....

[oshea.emma]

Hi there guys
How's things?
I'm working on trying to understand this question. Would really like a bit of direction.

Question
Given two real numbers a and b where a>1 and b>1 prove that

1/(logb a) + 1/(loga b)>= (greater or equal to) 2.

So far i got
Using rule for base change:
=loga b + 1/(loga b) >= 2
=(loga b)^2 +1 >= 2
=(loga b)^2 -1>=0

this is where i'm a little stuck.
Is it ok to say that let x=loga b
so we get
x^2 -1>>=0
=x^2-1^2>=0 ............... am i allowed to do this? because -1^2 =1 and not -1
=(x-1)^2>=0
This last line would work if a and b are greater then 1. Correct?

THANKS FOR THE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[pka]

Have you noticed the following?
\(\displaystyle \L
\begin{array}{l}
\left. \begin{array}{l}
\log _a (b) = x\quad \Rightarrow \quad a^x = b \\
\log _b (a) = y\quad \Rightarrow \quad b^y = a \\
\end{array} \right\} \Rightarrow \quad b^{xy} = a^x = b \Rightarrow \quad x = \frac{1}{y} \\
\end{array}\).

If \(\displaystyle \L
y > 0\quad \Rightarrow \quad y + \frac{1}{y} \ge 2\quad \& \quad y = \frac{1}{x}\).

 

[oshea.emma]

Thanks for the reply!!
Elegant answer,
how did you manage to get b^xy =a^x?

Does my attempt make sense to you?
Thanks again

 

[pka]

1) Does my attempt make sense to you?
2) How did you manage to get b^xy =a^x?

On 1). Frankly, I just cannot follow your notation sorry to say.
Maybe you could try to learn LaTeX.

For 2), we know that \(\displaystyle b^y = a\), so apply the power of x to both sides \(\displaystyle b^{xy} = a^x\).

Because \(\displaystyle b^{xy} = b\) the xy exponent must be 1.

 

[galactus]

Hi there guys
How's things?
I'm working on trying to understand this question. Would really like a bit of direction.

Question
Given two real numbers a and b where a>1 and b>1 prove that

1/(logb a) + 1/(loga b)>= (greater or equal to) 2.

So far i got
Using rule for base change:
=loga b + 1/(loga b) >= 2
=(loga b)^2 +1 >= 2
=(loga b)^2 -1>=0

this is where i'm a little stuck.
Is it ok to say that let x=loga b
so we get
x^2 -1>>=0
=x^2-1^2>=0 ............... am i allowed to do this? because -1^2 =1 and not -1
=(x-1)^2>=0
This last line would work if a and b are greater then 1. Correct?

THANKS FOR THE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I will try to interpret.

\(\displaystyle \L\\\frac{1}{log_{b}a}+\frac{1}{log_{a}b}\geq{2}\)

Provided a>1 and b>1.

So far i got
Using rule for base change:
\(\displaystyle =log_{a}b+\frac{1}{log_{a}b}\geq{2}\\
=(log_{a}b)^{2}+1\geq{2}\\
=(log_{a} b)^{2}-1\geq{0}\)

this is where i'm a little stuck.
Is it ok to say that let \(\displaystyle x=log_{a}b\)
so we get
\(\displaystyle x^{2} -1\geq{0}\)
\(\displaystyle =x^{2}-1^{2}\geq{0}\) ............... am i allowed to do this? because \(\displaystyle {-1^{2}} =1\)
\(\displaystyle =(x-1)^{2}\geq{0}\)
This last line would work if a and b are greater then 1. Correct?

Is this what you meant?. If you have time, learn some LaTex if you plan on posting to any extent. Click on quote at the upper right of the post to see some of the code used in this post.

 

[oshea.emma]

Yes! That's my question done in latex form Thank you very much!! I will try and learn it....so much to learn so little time.

So any ideas on the part that i'm stuck on?? Thanks for your time and efforts!! Really appreciated!

 

[pka]

Clearly it is true that: \(\displaystyle \log _a (b) + \frac{1}{{\log _a (b)}} \ge 2\).

From which it follows that: \(\displaystyle \left( {\log _a (b)} \right)^2 + 1 \ge 2\left( {\log _a (b)} \right)\).

However, it does not follow that: \(\displaystyle \left( {\log _a (b)} \right)^2 + 1 \ge 2\).

For example let a=2 and b=1.1 .