Question

Find the value of \[\log {10^a} + \log {10^{a + b}} + \log {10^{a + 3b}} + ....n\]

Answer 1

Hint: Radicand of logarithmic term under addition gets multiplied then apply formula of natural series to get result. Thereafter, we put the value of \[log10 = 1\]

Formula used: \[log{a^m} = m\log a,\,\,\log 10 = 1\]
\[1 + 2 + 3 + ...n = \dfrac{{n(n + 1)}}{2}\](Natural series)

Complete step by step answer:
(1) We have \[\log {10^a} + \log {10^{a + b}} + \log {10^{a + 2b}} + \log {10^{a + 3b}} + ....n\] terms.
Using logarithmic formula,
\[\log {a^m} = m\log a\]
We have, \[(a)\log 10 + (a + b)\log 10 + (a + 2b)\log 10 + ....n\]
(3) Also we know that \[\log 10 = 1\], use this in the above term
$a \times 1 + (a + b) \times 1 + (a + 2b) \times 1 + (a + 3b) \times 1,...... + n$
\[ \Rightarrow \,\,a + (a + b) + (a + 2b) + (a + 3b) + ....n\]
(4) Separating series as \[(a + a + a + .....n) + (b + 2b + 3b + ....n)\]
(5) \[na + b(1 + 2 + 3 + ....n)\] \[ \ldots ..\left( 1 \right)\]
(6) Using formula of natural series
\[1 + 2 + 3 + ......n = \dfrac{{n(n + 1)}}{2}\]
From equation (1)
(7) \[na + b\dfrac{{n(n + 1)}}{2}\]

Hence, the required value of the given series is \[na + \dfrac{{bn(n + 1)}}{2}\].

Note: In logarithmic series, powers of radicand become multipliers. A logarithm tells what exponent (or power) is needed to make a certain number, so logarithms are the inverse (opposite) of exponentiation.