Question

If $3 = k{2^r}$ and $15 = k{4^r}$ then the value of r will be
A.${\log _2}5$
B. 2
C.${\log _2}10$
D.4

Answer 1

We are given with two equations and both having the letter r in index form . If we observe that we have 2 in one term and 4 in another term and this is our hint to lead towards the solution. Now we will put 4 as a square of 2 and first find the value of k and then putting this value of k in any equation given we will find the value of r. Simple!

Complete step-by-step answer:
We are given that
$3 = k{2^r}..... \to equation1$
Also
$15 = k{4^r}..... \to equation2$
Now we can write 4 as a square of 2.
$\Rightarrow 15 = k{\left( {{2^2}} \right)^r}$
Using laws of indices ${\left( {{a^m}} \right)^n} = {\left( {{a^n}} \right)^m}$ above equation changes to
$\Rightarrow 15 = k{\left( {{2^r}} \right)^2}$
Now from equation1 we can write ${2^r} = \dfrac{3}{k}$
So this changes the equation to
$\Rightarrow 15 = k{\left( {\dfrac{3}{k}} \right)^2}$
Now taking the square we get,
$\Rightarrow 15 = k\left( {\dfrac{9}{{{k^2}}}} \right)$
Cancelling k from denominator with the one in numerator we get,
$\Rightarrow 15 = \dfrac{9}{k}$
Rearranging the terms
$\Rightarrow k = \dfrac{9}{{15}}$
Simplifying the terms
$\Rightarrow k = \dfrac{3}{5}$
This is the value of k. Now putting it in any of the equations above (we will put it in equation1). Thus it changes to,
$3 = \dfrac{3}{5}{2^r}..... \to equation1$
Cancelling 3 from both sides we get
$5 = {2^r}$
Applying log to both sides,
$\Rightarrow \log 5 = \log {2^r}$
Now we know that $\log {a^m} = m\log a$ thus above equation is simplified as
$\Rightarrow \log 5 = r\log 2$
To find the value of r let’s take logs terms on one side
$\Rightarrow \dfrac{{\log 5}}{{\log 2}} = r$
$\Rightarrow {\log _2}5 = r..... \to \dfrac{{\log a}}{{\log b}} = {\log _b}a$
Hence we get the value of r as $\Rightarrow r = {\log _2}5$
So option A is correct.

Note: Alternative method:
Students can also take the ratio of equation1 and equation2. This will save your time and we need not to find the value of k.
$\Rightarrow \dfrac{3}{{15}} = \dfrac{{k{2^r}}}{{k{4^r}}}$
Cancelling k

$$\Rightarrow \dfrac{1}{5} = \dfrac{{{2^r}}}{{{4^r}}} \\\ \Rightarrow \dfrac{1}{5} = \dfrac{{{2^r}}}{{{{({2^r})}^2}}} \\\ \Rightarrow \dfrac{1}{5} = \dfrac{1}{{{2^r}}} \\\ \Rightarrow {2^r} = 5 \\\$$

Now we came to same step of applying log and the answer is $\Rightarrow r = {\log _2}5$