Question

The value ${(0.16)^{{{\log }_{2.5}}\\{ \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + .....\\} }}$ is
a. 2
b. 4
c. 6
d. 8

Answer 1

Start by solving each step one by one. First, we solve the infinite GP by using the formula sum of a G.P which is $= \dfrac{a}{{1 - r}}$, then we’ll proceed toward the logarithm function, we’ll simplify it using the properties of the logarithm ${\text{alog(x) = log(}}{{\text{x}}^{\text{a}}}{\text{)}}$ and ${{\text{a}}^{{\text{lo}}{{\text{g}}_{\text{a}}}{\text{b}}}}{\text{ = b}}$, to get the required answer.

Complete step by step Answer:

We have the sum of the infinite GP, where a is the initial term and r is the common ratio$= \dfrac{{\text{a}}}{{{\text{1 - r}}}}$given that $|r| < 1$ .
So, we have, $\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + .......$
The common ratio(r) will the ratio any term with its preceding term
i.e. $r = \dfrac{{\dfrac{1}{{{3^2}}}}}{{\dfrac{1}{3}}}$
$\therefore r = \dfrac{1}{3}$
So we have an infinite G.P. where a is$\dfrac{1}{3}$ and r is $\dfrac{1}{3}$
Hence substituting the value of a and r in the formula of the sum of infinite GP which is $= \dfrac{a}{{1 - r}}$, we get,
$= \dfrac{{\dfrac{1}{3}}}{{(1 - \dfrac{1}{3})}}$
$= \dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}$
$= \dfrac{1}{2}$
We know that 2.5 $= \dfrac{5}{2}$and $0.16 = \dfrac{4}{{25}}$
We get, ${(0.16)^{{{\log }_{2.5}}\\{ \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + .....\\} }}$
= ${(\dfrac{4}{{25}})^{{{\log }_{\dfrac{5}{2}}}\dfrac{1}{2}}}$
We can write $\dfrac{1}{2} = {2^{ - 1}}$
We know that, ${(\dfrac{5}{2})^{ - 2}} = {\dfrac{4}{{25}}^{}}$
now, ${(\dfrac{4}{{25}})^{{{\log }_{\dfrac{5}{2}}}\dfrac{1}{2}}} = {(\dfrac{5}{2})^{ - 2{{\log }_{\dfrac{5}{2}}}\dfrac{1}{2}}}$
Now we use the fact ${\text{alog(x) = log(}}{{\text{x}}^{\text{a}}}{\text{)}}$, we get,
$= {(\dfrac{5}{2})^{{{\log }_{\dfrac{5}{2}}}{{(\dfrac{1}{2})}^{ - 2}}}}$
$= {(\dfrac{5}{2})^{{{\log }_{\dfrac{5}{2}}}4}}$
As, ${{\text{a}}^{{\text{lo}}{{\text{g}}_{\text{a}}}{\text{b}}}}{\text{ = b}}$we see,${(\dfrac{5}{2})^{{{\log }_{\dfrac{5}{2}}}4}} = 4$, which is an option (b)

Note: Always first simplify the question, we can take different parts of the question. The formulas we had used in the problem are, ${{\text{a}}^{{\text{lo}}{{\text{g}}_{\text{a}}}{\text{b}}}}{\text{ = b}}$, ${\text{alog(x) = log(}}{{\text{x}}^{\text{a}}}{\text{)}}$and the sum of the infinite GP$= \dfrac{{\text{a}}}{{{\text{1 - r}}}}$.
Here so we used the sum of G.P. $S = \dfrac{a}{{1 - r}}$as$r = \dfrac{1}{3}$ but if$\left| r \right| \geqslant 1$ then we would use the formula for sum of G.P. as${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$