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  • Let a, b, c be three distinct positive real numbers such that <img alt="(2 a)^{\log _{g_c} a}=(b c)^{\log _c b} \text { and } b^{\log _c 2}=a^{\log _c c} \text {. }" src="https://entrancecorne

Let a, b, c be three distinct positive real numbers such that (2 a)^{\log _{g_c} a}=(b c)^{\log _c b} \text { and } b^{\log _c 2}=a^{\log _c c} \text {. }

Then 6a + 5bc is equal to _____.

Option: 1

Bouns


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

(2 a)^{\ln a}=(b c)^{\ln b} \quad 2 a>0, b c>0

{\ln a}(ln2+Ina)={\ln b}(ln\, b+ln\, c)\\

\ln 2 \cdot \ln b=\ln c \cdot \ln a

\ln 2 =\alpha ,\, \ln a=x,\ln b=y,\ln c=z

\alpha y=xz

\begin{aligned} & x(\alpha+x)=y(y+z) \\ & \alpha=\frac{x z}{y} \\ & x\left(\frac{x z}{y}+x\right)=y(y+z) \\ & x^2(z+y)=y^2(y+z) \\ & y+z=0 \text { or } x^2=y^2 \Rightarrow x=-y \end{aligned}

bc=1 \: \: or\: \: ab=1

bc=1 \: \: or\: \: ab=1

(\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\frac{1}{2}, \lambda, \frac{1}{\lambda}\right), \lambda \neq 1,2, \frac{1}{2}

then

6a+5bc=3+5=8\\

\text { (II) }(\mathrm{a}, \mathrm{b}, \mathrm{c})=\left(\lambda, \frac{1}{\lambda}, \frac{1}{2}\right), \lambda \neq 1,2, \frac{1}{2}

In this situation infinite answer are possible
So, Bonus.

 

Posted by

HARSH KANKARIA

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