Question

If a=log12 base 24,b=log24 base 36,c=log 36 base 48,Then show that 1+abc=2bc

Answer 1

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Answer 2

Answer:

Given, $a=\log _{24} 12, b=\log _{36} 24, c=\log _{48} 36$

Required to Prove – 1 + abc = 2bc

            LHS       RHS  

LHS- 1 + abc =  $\begin{array}{l}{1+\left(\log _{24} 12\right)\left(\log _{36} 24\right)\left(\log _{48} 36\right)} \\ {=1+\left(\frac{\log 12}{\log 24}\right)\left(\frac{\log 24}{\log 36}\right)\left(\frac{\log 36}{\log 48}\right)} \\ {=1+\left(\frac{\log 12}{\log 48}\right)}\end{array}$

$\begin{aligned} &=\left(\frac{\log 48+\log 12}{\log 48}\right) \\ &=\frac{\log (48 \times 12)}{\log 48} \\ &=\frac{\log 576}{\log 48} \end{aligned}$

$\begin{array}{l}{=\frac{\log (24)^{2}}{\log 48}} \\ {=\frac{2 \log 24}{\log 48}} \\ {=2 \times\left(\frac{\log 24}{\log 36}\right)\left(\frac{\log 36}{\log 49}\right)}\end{array}$

$\begin{aligned} &=2 \times\left(\log _{36} 24\right)\left(\log _{48} 36\right) \\ &=\bold{2 \mathrm{bc}} \end{aligned}$

Hence proved