$$\mbox{If}\ \log_{2a}\left(a\right) = x,\ \log_{3a}\left(2a\right) = y,\ \log_{4a}\left(3a\right) = z.\quad \mbox{Then, what is the value of}\ xyz-2yz\,?. $$ Not exactly able to solve it any further.
Asked
Active
Viewed 75 times
0
-
Do you know about change of base identity? – Saket Choudhary Jun 27 '16 at 17:27
1 Answers
1
Hint. One may write $$ x=\log_{2a}a=\frac{\log(a)}{\log(2a)},\quad y=\log_{3a}(2a)=\frac{\log(2a)}{\log(3a)},\quad z=\log_{4a}(3a)=\frac{\log(3a)}{\log(4a)} $$ giving $$ xyz-2yz=\frac{\log(a)}{\log(2a)}\cdot\frac{\log(2a)}{\log(3a)}\cdot\frac{\log(3a)}{\log(4a)}-2\cdot\frac{\log(2a)}{\log(3a)}\cdot\frac{\log(3a)}{\log(4a)} $$ that is
$$ xyz-2yz=\frac{\log(a)}{\log(4a)}-\frac{2 \cdot \log(2a)}{\log(4a)}. $$
Can you take it from here?
Olivier Oloa
- 120,989
-
Thanks for the answer but can you please solve it further I'm a little confused. – Ujjwal Jun 27 '16 at 17:32
-
-
-
-
-
-
I got the answer just one last question..i know that loga+logb = loga. b but what is loga-logb=?? Is it -loga.b? – Ujjwal Jun 27 '16 at 17:41
-
Finally $xyz-2yz=\frac{\log(a)}{\log(4a)}-\frac{2 \cdot \log(2a)}{\log(4a)}=\frac{-\log(4a)}{\log(4a)}=-1.$ – Olivier Oloa Jun 27 '16 at 17:41
-
-
-
i know that loga+logb = loga. b but what is loga-logb=?? Is it -loga.b? – Ujjwal Jun 27 '16 at 17:44
-
-
After taking LCM shouldn't it be like $\frac{log a - 2(log a+ log 2)}{log 4a}$ – Ujjwal Jun 27 '16 at 17:49
-