Given that $A=\text{log}_{16}15$ and $B=\text{log}_{12}18$, find $\text{log}_{25}24$ in terms of A and B.

Question

Given that $A=\text{log}_{16}15$ and $B=\text{log}_{12}18$, find $\text{log}_{25}24$ in terms of $A$ and $B$.

I found that the answer is $\frac{B-5}{-2AB-2A+4B-2}$ but I used a very inefficient steps to get this answer (not sure correct or not). My approach involved obtaining the highest common factor as a base (for log), but the solution turned out to be excessively lengthy. Is there a efficient approach?

Answer 1

Try using only logarithms with base 5 (transforming A and B in these forms is fairly simple using logarithms properties). You will end up with expressions in which the only logarithms are base 5 with 2 o 3 as arguments; from there it shouldn' be hard as the value you're after is $1/2(3\text{log}_5 2+\text{log}_5 3)$

Answer 2

I would want to express everything in terms of $x=\log_2 3$ and $y=\log_2 5$.

\begin{align}\log_{24}{25}&=\frac{\log_2{24}}{\log_2{25}}\\ &=\frac{\log_28+\log_23}{\log_25^2} \\&=\frac{3+x}{2y}\\\hline\end{align}

\begin{align} A&= \log_{2^4}(3\times 5)\\ &=\frac{x+y}{4}\\\hline\end{align}

\begin{align} B&=\log_{12}\frac{12^2}{2^3} \\&=2-3\log_{12}2\\&= 2-\frac{3}{\log_2(2^2\times 3)}\\ &=2-\frac{3}{2+x}\\ \implies x&=\frac{3}{2-B}-2\end{align}

Now, "solve" for $y$ as well and back-substitute.

I don't know if that's shorter than your approach...

Answer 3

Let $~\displaystyle C = \log_{25} (24).~$
The goal is to express $~C~$ in terms of $~A~$ and $~B.~$

My approach will be to express $~\log 3~$ in terms of $~B~$ and $~\log 2.~$ Then, I will express $~\log 5~$ in terms of $~A, ~B, ~$ and $~\log 2.~$ Then, I will express $~C~$ in terms of $~A~$ and $~B.~$

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$$B = \log_{12}18 = \frac{\log 18}{\log 12} = \frac{\log 2 + 2\log 3}{2\log 2 + \log 3} \implies $$

$$B (2\log 2 + \log 3) = \log 2 + 2\log 3 \implies $$

$$\log 3 (B - 2) = \log 2 (1 - 2B) \implies $$

$$\log 3 = \log 2 \left[ ~\frac{1 - 2B}{B - 2} ~\right]. \tag1 $$

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$$A = \frac{\log 15}{\log 16} = \frac{\log 3 + \log 5}{4\log 2} \implies $$

$$\log 5 = 4A\log 2 - \log 3$$ $$= \log 2 ~\left[ ~4A - \frac{1 - 2B}{B - 2} ~\right] = \log 2 ~\frac{4AB - 8A - 1 + 2B}{B-2}. \tag2 $$

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$$C = \frac{\log 24}{\log 25} = \frac{3\log 2 + \log 3}{2 \log 5}$$

$$= \frac{3\log 2 + \log 2\left[ ~\frac{1 - 2B}{B - 2} ~\right] }{2\log 2 ~\frac{4AB - 8A - 1 + 2B}{B-2} }.$$

Dividing both the numerator and denominator by $~\dfrac{\log 2}{B-2}~$ gives

$$C = \frac{3(B-2) + (1 - 2B)}{2(4AB - 8A - 1 + 2B)} = \frac{B-5}{8AB - 16A - 2 + 4B}.$$

Answer 4

Beneficial to use log formula

$$ \log_b~a=\frac{\log a}{\log b}$$

for any base. Expand for $A,B$ in terms of $(\log2, \log3); $

Solving two simultaneous equations for $(\log2, \log3) $

$$\log 2= ((-2 + B) \log5)/(-1 - 8 A + 2 B + 4 A B)$$

$$ \log 3= -((-1 + 2 B) \log5)/(-1 - 8 A + 2 B + 4 A B)$$

Plugging in I find the answer:

$$\dfrac{B-5}{4B -16A+8AB-2} . $$

(Took help of Mathematica as follows).
Solve[{(LG5 + three) /(4 two) == A, (two + 2 three) /(three + 2 two) ==  B}, {two, three}]
two1 = ((-2 + B) LG5)/(-1 - 8 A + 2 B + 4 A B); 
three1 = -(((-1 + 2 B) LG5)/(-1 - 8 A + 2 B + 4 A B)); Simplify[(three1 + 3 two1)/(2 LG5)]

Answer 5

You have $3$ logarithms involved $\ln(2),\ln(3),\ln(5)$ for only $2$ variables $A,B$ so you somehow have to trust the problem that a simplification will occur.

$\ln(5)$ only appear in $A=\dfrac{\ln(15)}{\ln(16)}=\dfrac{\ln(3)+\ln(5)}{4\ln(2)}$

so I want to express it in term of $A$.

$$\ln(5)=4A\ln(2)-\ln(3)$$

Then I'd like to express for instance $\ln(3)$ in term of $B=\dfrac{\ln(18)}{\ln(12)}=\dfrac{2\ln(3)+\ln(2)}{2\ln(2)+\ln(3)}$.

$$\ln(3)=\left(\dfrac{2B-1}{2-B}\right)\ln(2)$$

Now notice that $\ln(2)$ will cancel in the resulting expression

$\require{cancel}C=\dfrac{\ln(24)}{\ln(25)}=\dfrac{3\ln(2)+\ln(3)}{2\ln(5)}=\dfrac{3\cancel{\ln(2)}+\frac{2B-1}{2-B}\cancel{\ln(2)}}{8A\cancel{\ln(2)}-2\frac{2B-1}{2-B}\cancel{\ln(2)}}=\dfrac{3(2-B)+2B-1}{8A(2-B)-4B+2}=\dfrac{5-B}{16A-8AB-4B+2}$