3
$\begingroup$

If it is given $$\log_ac=\log_{\sqrt{a}}b+\log_ab^2+\log_{a^2}b^3+\log_{a^4}b^4+\log_{a^8}b^5+\cdots$$ and $a,b,c>1$ then find the least value of $$\log_a{bc}+\log_bac+\log_cab$$

Rewriting the first equation after doing some manipulation gives $$\log_ac=\log_{\sqrt{a}}b+\sum_{i=0}^{\infty}\log_{a^{2^i}}b^{i+2}$$ $$\implies \log_ac=2\log_{a}b+\log_ab\cdot\sum_{i=0}^{\infty}\frac{i+2}{2^i}$$ $$\implies \log_ac=2\log_{a}b+6\log_ab$$ $$\implies \log_ac=8\log_{a}b$$ $$\implies c=b^8$$ Now, $$\log_a{bc}+\log_bac+\log_cab=\log_ab+\log_ac+\log_ba+\log_bc+\log_ca+\log_cb$$ $$=\log_ab+\log_ac+\frac{1}{\log_ab}+\frac{1}{\log_ac}+8+\frac18$$ $$\ge2+2+8+\frac{1}{8}=\boxed{12.125}$$

However, I don't think that it is the right answer as in the original question it was asked to find the value of $w+x+y+z$ where least value was equal to $\displaystyle\frac{w+x\sqrt{y}}{z}$ where $w,x,z$ are relatively prime and $y$ is not a square of any prime.

Am I right or wrong. Guide me if I'm wrong. Any help is greatly appreciated.

$\endgroup$
2
  • $\begingroup$ Are $a,b,c$ all integers? And are $w,x,y,z$ (and thus necessarily $\frac{w+x\sqrt y}z$) integers? $\endgroup$ Commented Feb 20, 2023 at 11:20
  • 1
    $\begingroup$ @student91 's answer is the best. Just thought I'd tell you that $\log_ab+\log_ac+\frac{1}{\log_ab}+\frac{1}{\log_ac}$ has a minimum of $2+2$ when $a=b=c$, which is a contradiction. That's why you're not getting the square root $\endgroup$ Commented Feb 20, 2023 at 11:53

2 Answers 2

Reset to default
3
$\begingroup$

I think your general approach is correct, but the bound could have been a bit tighter:

When you have $\log_ab+\log_ac+\frac1{\log_ab}+\frac1{\log_ac}+8+\frac18$, you can use $c=b^8$ to obtain $$\log_ab+\log_ac+\frac1{\log_ab}+\frac1{\log_ac}+8+\frac18=\log_ab+8\log_ab+\frac1{\log_ab}+\frac1{8\log_ab}+8+\frac18$$$$=9\log_ab+\frac9{8\log_ab}+\frac{65}8.$$

Now $x+\frac1{8x}$ has minimum $\frac1{\sqrt2}$ so we obtain $$9\log_ab+\frac9{8\log_ab}+\frac{65}8\ge \frac9{\sqrt2}+\frac{65}8\gtrapprox14.48896$$

$\endgroup$
1
  • $\begingroup$ you beat me to it(+1) $\endgroup$ Commented Feb 20, 2023 at 11:49
0
$\begingroup$

$$\log_{\sqrt a}b=\frac{\ln b}{(1/2)\ln a}=2\log_ab$$ $$\implies\log_ac=2\log_ab+\log_ab^2=4\log_ab\implies c=b^4$$ The expression being minimised then actually simplifies to $5\log_ab+\frac54\log_ba+\frac{17}4$, and so the minimum of $x+\frac1x$ being $2$ does not apply here. Let $x=\log_ab$, then the minimum of $x+\frac1{4x}$ is $1$ at $x=\frac12$, so the expression's minimum is $5+\frac{17}4=\frac{37}4$.

$\endgroup$
1
  • 1
    $\begingroup$ how did you get $c=b^4$ instead of $c=b^8$ what OP obtained? $\endgroup$ Commented Feb 20, 2023 at 11:31

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.