Logarithmic function

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0 replies

jlacosta

May 1, 2025

0 replies

Challenge Problem: triangle inequality

Bottema 7

N 24 minutes ago by Speedysolver1

Prove that in any triangle we have:

a^{2}+b^{2}+c^{2} \geq 4sqrt{3}S

7 replies

Bottema

May 12, 2004

Speedysolver1

24 minutes ago

Pertenacious Polynomial Problem

BadAtCompetitionMath21420 3

N 24 minutes ago by pieMax2713

Let the polynomial $P(x) = x^3-x^2+px-q$ have real roots and real coefficients with $q>0$ . What is the maximum value of $p+q$ ?

This is a problem I made for my math competition, and I wanted to see if someone would double-check my work (No Mike allowed):

solution

We need to maximize $p+q$ because we know $q > 0$ , then $p>0$ . Let $P$ have roots $a,b,c$ . Observe that $a+b+c>0, ab+bc+ca, abc>0$ . Obscure, yes, but this means $a,b,c>0$ (I won't be proving this here). This new information allows us to utilize the mean inequality chain as follows:

By QM-AM,
$\begin{align*} \sqrt{\frac{(a+b+c)^2 - 2(ab+cb+ca)}{3}} & \geq \frac{a+b+c}{3},\\ 1-2p & \geq \frac{1}{3},\\ p \leq \frac{1}{3} \end{align*}$ Similarly, applying AM-HM yields $q\leq 1/27$ . Thus, $p+q \leq10/27$ .

Is this solution good?

3 replies

BadAtCompetitionMath21420

Today at 3:13 AM

pieMax2713

24 minutes ago

2022 MARBLE - Mock ARML I -8 \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=32

parmenides51 2

N 2 hours ago by Kempu33334

Let $a,b,c$ complex numbers with $ab+ +bc+ca = 61$ such that
$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}= 5$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=32.$ Find the value of $abc$ .

2 replies

parmenides51

Jan 14, 2024

Kempu33334

2 hours ago

Inequalities

sqing 13

N 3 hours ago by sqing

Let $a,b,c>0 , a+b+c +abc=4$ . Prove that
$\frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$ Let $a,b,c>0 , ab+bc+ca+abc=4$ . Prove that
$\frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$

13 replies

1 viewing

sqing

May 15, 2025

sqing

3 hours ago

No more topics!

Logarithmic function

jonny 2

N May 11, 2025 by KSH31415

If $\log_{6}(15) = a$ and $\log_{12}(18)=b,$ Then $\log_{25}(24)$ in terms of $a$ and $b$

2 replies

jonny

Jul 15, 2016

KSH31415

May 11, 2025

Logarithmic function

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jonny

3455 posts

jonny

#1 h Jul 15, 2016, 12:33 PM • 1 Y

Y by Adventure10

If $\log_{6}(15) = a$ and $\log_{12}(18)=b,$ Then $\log_{25}(24)$ in terms of $a$ and $b$

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Mathzeus1024

895 posts

Mathzeus1024

#2 h May 11, 2025, 2:26 PM

Y by

We have:

$a = \frac{\ln(15)}{\ln(6)} = \frac{\ln(3)+\ln(5)}{\ln(3)+\ln(2)}$ (i);

$b = \frac{\ln(18)}{\ln(12)} = \frac{2\ln(3)+\ln(2)}{\ln(3)+2\ln(2)}$ (ii).

If $\frac{\ln(24)}{\ln(25)} = \frac{3\ln(2)+\ln(3)}{2\ln(5)}$ , then we obtain:

$\frac{\ln(24)}{\ln(25)}= \frac{\ln(3)+2\ln(2)+\ln(2)}{2\ln(5)} = \frac{\frac{2\ln(3)+\ln(2)}{b}+\ln(2)}{2[a(\ln(2)+\ln(3))-\ln(3)]} = \textcolor{red}{\frac{\ln(18)+b\ln(2)}{2b[a\ln(6)-\ln(3)]}}$ .

This post has been edited 1 time. Last edited by Mathzeus1024, May 11, 2025, 2:29 PM

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KSH31415

398 posts

KSH31415

#3 h May 11, 2025, 11:09 PM

Y by

Mathzeus1024 wrote:

$\textcolor{red}{\frac{\ln(18)+b\ln(2)}{2b[a\ln(6)-\ln(3)]}}$ .

I would assume that the intent of the problem is to express $\log_{25}24$ with $a$ and $b$ an no other logarithms. This is what I did.

I'm going to brush over some of the log identities and manipulations. If anybody wants me to explain, I can. Let $x=\log_52$ and $y=\log_53$ . Then
$a=\frac{1+ y}{x+y}$ and
$b=\frac{2y+x}{2x+y}.$ We'll find $x$ and $y$ in terms of $a$ and $b$ , which will then give us
$\log_{25}24=\frac{3x+y}2.$ Rearranging, we get
$ax+ay=y+1\implies ax=y(1-a)+1~~~(1)$ and
$2bx+by=2y+x\implies (2b-1)x=(2-b)y\implies y=\frac{2b-1}{2-b}x. ~~~(2)$ Plugging $(2)$ into $(1)$ , we get
$ax=\frac{(2b-1)(1-a)}{2-b}x+1=\frac{2b-1-2ab+a}{2-b}x+1\implies \frac{2a-2ab-(2b-1-2ab+a)}{2-b}x=\frac{a-2b+1}{2-b}x=1\implies x=\frac{2-b}{a-2b+1}.$ Then $y=\frac{2b-1}{a-2b+1}$ , giving
$\log_{25}24=\frac{3x+y}2=\frac{6-3b+4b-2}{2(a-2b+1)}=\boxed{\frac{b+4}{2a-4b+2}}.$

There is probably an algebra mistake in there somewhere.

Surely there is a more direct way to find this (though my answer doesn't hint at any). Can anybody find it?

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