1
$\begingroup$

So, I've tried using the properties:

$$\log_a b = \frac{\log_c b}{\log_c a}$$

and..

$$\log_a bc = \log_a b + \log_a c$$

And, the final simplification should be in the following options:

$$A. \frac{1-a+abc}{2a(1+b)}$$ $$B. \frac{1-a-abc}{2a(1+b)}$$ $$C. \frac{1+a-abc}{2a(1+b)}$$ $$D. \frac{1+a+abc}{2a(1+b)}$$

I also tried substituting values in the options but it takes eternity to finish.... 😅😅

Plz help....

$\endgroup$
2
  • 1
    $\begingroup$ Your change of base formula is incorrect. The correct formula is $\log_a b = \frac{\log_c b}{\log_c a}$. $\endgroup$ Commented May 15 at 17:08
  • $\begingroup$ Ohh! srry 😅, Lemme correct it lol $\endgroup$
    – Mune
    Commented May 15 at 17:11

5 Answers 5

3
$\begingroup$

$$\begin{align*} \log_{225}70&=\log_{225}(2\times 5\times 7)=\log_{225}2+\log_{225}5+\log_{225}7\\ &=\dfrac{1}{\log_2 225}+\dfrac{1}{\log_5 225}+\dfrac{1}{\log_7 225}\\ &=\dfrac{1}{2\log_2 3+2\log_2 5}+\dfrac{1}{2\log_5 3+2}+\dfrac{1}{2\log_7 3+2\log_7 5}\\ &=\dfrac{1}{2/c+2/bc}+\dfrac{1}{2+2b}+\dfrac{1}{2ab+2a}\\ &=\dfrac{1+a+abc}{2a(b+1)} \end{align*}$$

$\endgroup$
1
  • $\begingroup$ I think, numerator is not correct. $\endgroup$
    – Bob Dobbs
    Commented May 15 at 17:37
2
$\begingroup$

So we could rewrite and create new logarithm terms with respect to a,b and c. In the question we need $$\log_{225} 70 .$$ Here 225 can be written as 3x3x5x5. Amd 70 can be written as 2×5×7. Now we get $$\frac{\log_2 2 + \log_2 5 +\log_2 7}{2(\log_2 3 + \log_2 5 )}$$ Now we can rewrite this as $$\frac{1+\frac{1}{bc}+\frac{1}{abc}}{2(\frac{1}{c}+\frac{1}{bc})}$$ After taking LCM we get $$\frac{1+abc+a}{2a(1+b)}$$ Hope this clears your doubt.

$\endgroup$
1
$\begingroup$

As an alternative way, using exponential, we have that

  • $\log_7 5 = a \iff 7^a=5$
  • $\log_5 3 = b \iff 5^b=3$
  • $\log_3 2 = c\iff 3^c=2$

then $$\log_{225}70 =x \iff 225^x=70 \iff (3^2\cdot 5^2)^x=(2 \cdot 5 \cdot 7)\iff 3^{2x\left(1+\frac1b\right)}=3^{c+\frac1b+\frac1{ab}}$$

therefore

$$x= \frac{c+\frac1b+\frac1{ab}}{2+\frac2b}=\frac{1+a+abc}{2a+2ab}$$

$\endgroup$
2
  • 1
    $\begingroup$ I was going to post something similar - if confused about logs and the base change formula it is always possible to go to powers. $\endgroup$ Commented May 15 at 20:01
  • $\begingroup$ Yes indeed this is a nice alternative! Thanks, Bye $\endgroup$
    – user
    Commented May 15 at 20:04
0
$\begingroup$

$$S= \log_{225}70=\frac{\ln70}{\ln225}=\frac{\ln2+\ln5+\ln7}{2(\ln3+\ln5)} $$ $$\log_53=b\implies \ln3=b\ln5$$ $$\log_32=c\implies\ln2=c\ln3=bc\ln5$$ $$\log_75=a\implies\ln7=\frac1a\ln5$$ So, $$S=\frac{bc+1+\frac1a}{2(b+1)}=\frac{1+a+abc}{2a(1+b)}$$ D

$\endgroup$
0
$\begingroup$

By the change of base formula, the given logarithms are equivalent to:

$$a = \frac{\ln 5}{\ln 7} \implies \ln 5 = a \ln 7 \tag{1}$$ $$b = \frac{\ln 3}{\ln 5} \implies \ln 3 = b \ln 5 \tag{2}$$ $$c = \frac{\ln 2}{\ln 3} \implies \ln 2 = c \ln 3 \tag{3}$$

Substituting (1) into (2), we get:

$$\ln 3 = ab \ln 7 \tag{4}$$

Substituting (2) into (3), we get:

$$\ln 2 = bc \ln 5 \tag{5}$$

And substituting (1) into (5) gives:

$$\ln 2 = abc \ln 7\tag{6}$$

So we can now write the three other logarithms in terms of $\ln 7$:

$$\ln 2 = abc \ln 7$$ $$\ln 3 = ab \ln 7$$ $$\ln 5 = a\ln 7$$

We are asked to find:

$$\log_{225} 70$$ $$=\frac{\ln 70}{\ln 225}$$ $$=\frac{\ln (2 \times 5 \times 7)}{\ln (3^2 \times 5^2)}$$ $$=\frac{\ln 2 + \ln 5 + \ln 7}{2\ln 3 + 2\ln 5}$$ $$=\frac{abc \ln 7 + a\ln 7 + \ln 7}{2ab \ln 7 + 2a\ln 7}$$ $$=\frac{abc + a + 1}{2ab + 2a}$$ $$=\frac{1 + a + abc}{2a(1 + b)}$$

Aka answer choice (D).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .