Question

$\bf\: 4 + log_2 (3x) = 10$​

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}\\ \\ \sf\: Given, \\ \\ \sf\implies\: 4 + log_{2} (3x) = 10 \\ \\ \sf\implies\: log_{2} (3x) = 10 - 4 \\ \\ \sf\implies\: log_{2} (3x) = 6 \\ \\ \underline {\sf { Use\: definition\: of\:Common\:Logarithm :}} \\ \\ \sf\implies\: 3x = 2^6 \\ \\ \sf\implies\: 3x = 64 \\ \\ \sf\implies\: x = \dfrac {64}{3} \\ \\ \underline {\sf {Formula \:used :}} \\ \\ \sf\: b^a = x \: if \: and \: only \: if \: log_b(x) = a$

Answer 2

Answer:

\begin{lgathered}\underline{\underline{\mathfrak{\Large{Solution : }}}}\\ \\ \sf\: Given, \\ \\ \sf\implies\: 4 + log_{2} (3x) = 10 \\ \\ \sf\implies\: log_{2} (3x) = 10 - 4 \\ \\ \sf\implies\: log_{2} (3x) = 6 \\ \\ \underline {\sf { Use\: definition\: of\:Common\:Logarithm :}} \\ \\ \sf\implies\: 3x = 2^6 \\ \\ \sf\implies\: 3x = 64 \\ \\ \sf\implies\: x = \dfrac {64}{3} \\ \\ \underline {\sf {Formula \:used :}} \\ \\ \sf\: b^a = x \: if \: and \: only \: if \: log_b(x) = a\end{lgathered}

Solution:

Given,

⟹4+log

2

(3x)=10

⟹log

2

(3x)=10−4

⟹log

2

(3x)=6

UsedefinitionofCommonLogarithm:

⟹3x=2

6

⟹3x=64

⟹x=

3

64

Formulaused:

b

a

=xifandonlyiflog

b

(x)=a