Question

Q1.a = log {12 base 24}, b = log {24 base 36}, c = log {36 base 48}, then 1 + abc is equal to?​

Answer 1

Step-by-step explanation:

Given :-

$\begin{gathered} \rm \rightarrow a = log_{24}(12) \\ \rm\rightarrow b = log_{36}(24) \\\rm\rightarrow c = log_{48}(36) \end{gathered} </p><p>$

To find :-

$\rm 1+abc</p><p>$

Solution :-

$\rm \longrightarrow 1+abc$

$\rm\longrightarrow1 + \bigg( log_{24}(12) \times log_{36}(24) \times log_{48}(36) \bigg)$

$\begin{gathered}\rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log(24)} \times \dfrac{log(24)}{log({36})} \times \dfrac{log(36)}{log({48})} \bigg) \\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{1} \times \dfrac{1}{1} \times \dfrac{1}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow 1 + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48}(48) + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48 }(48 \times 12) \\ \\ \\ \rm\longrightarrow log_{48}{576}\\ \\ \\ \rm\longrightarrow log_{48}{ {(24)}^{2} } = 2log_{48}{ {(24)} }\\ \\ \\ \rm\longrightarrow 2log_{48}{ {(24)} } \times log_{36}{ {(36)} }\\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{48}} \times \dfrac{log_{36}}{log_{36}} \\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{36}} \times \dfrac{log_{36}}{log_{48}} \\ \\ \\ \rm\longrightarrow 2 \times b \times c\\ \\ \\ \rm\longrightarrow 2bc\end{gathered}$

Answer :-

Answer :-The correct answer is option (c) 2bc

Answer 2

Answer:

Step-by-step explanation:

Given :-

\begin{gathered}\begin{gathered} \rm \rightarrow a = log_{24}(12) \\ \rm\rightarrow b = log_{36}(24) \\\rm\rightarrow c = log_{48}(36) \end{gathered} < /p > < p > \end{gathered}

→a=log

24

(12)

→b=log

36

(24)

→c=log

48

(36)

</p><p>

To find :-

\rm 1+abc < /p > < p >1+abc</p><p>

Solution :-

\rm \longrightarrow 1+abc⟶1+abc

\rm\longrightarrow1 + \bigg( log_{24}(12) \times log_{36}(24) \times log_{48}(36) \bigg)⟶1+(log

24

(12)×log

36

(24)×log

48

(36))

\begin{gathered}\begin{gathered}\rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log(24)} \times \dfrac{log(24)}{log({36})} \times \dfrac{log(36)}{log({48})} \bigg) \\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{1} \times \dfrac{1}{1} \times \dfrac{1}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow 1 + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48}(48) + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48 }(48 \times 12) \\ \\ \\ \rm\longrightarrow log_{48}{576}\\ \\ \\ \rm\longrightarrow log_{48}{ {(24)}^{2} } = 2log_{48}{ {(24)} }\\ \\ \\ \rm\longrightarrow 2log_{48}{ {(24)} } \times log_{36}{ {(36)} }\\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{48}} \times \dfrac{log_{36}}{log_{36}} \\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{36}} \times \dfrac{log_{36}}{log_{48}} \\ \\ \\ \rm\longrightarrow 2 \times b \times c\\ \\ \\ \rm\longrightarrow 2bc\end{gathered} \end{gathered}

⟶1+(

log(24)

log(12)

×

log(36)

log(24)

×

log(48)

log(36)

)

⟶1+(

1

log(12)

×

1

1

×

log(48)

1

)

⟶1+(

log(48)

log(12)

)

⟶1+log

48

(12)

⟶log

48

(48)+log

48

(12)

⟶log

48

(48×12)

⟶log

48

576

⟶log

48

(24)

2

=2log

48

(24)

⟶2log

48

(24)×log

36

(36)

⟶2×

log

48

log

24

×

log

36

log

36

⟶2×

log

36

log

24

×

log

48

log

36

⟶2×b×c

⟶2bc

Answer :-

Answer :-The correct answer is option (c) 2bc