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

Hint: 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${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$