Question 49

For a real number a, if $$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$ then a must lie in the range

Solution

We have :$$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$$
We get $$\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$$
we get $$\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$$
we get $$\left(\log32\ +\log\ 15\right)=4\log a$$
=$$\log480=\log a^4$$
=$$a^4\ =480$$
so we can say a is between 4 and 5 .

Video Solution

video

Create a FREE account and get:

  • All Quant CAT Formulas and shortcuts PDF
  • 33+ CAT previous papers with video solutions PDF
  • Topic-wise Previous year CAT Solved Questions for Free

CAT Quant Questions | CAT Quantitative Ability

CAT DILR Questions | LRDI Questions For CAT

CAT Verbal Ability Questions | VARC Questions For CAT

Related Formulas With Tests

cracku

Boost your Prep!

Download App