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

Hint: 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 \[ \Rightarrow r = {\log _2}5\]