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Logarithms

Section: Mathematics

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Question : 6 of 11

Marks: +1, -0

Let

a, b, c

be three distinct positive real numbers such that

(2a)^{\log_e a} = (bc)^{\log_e b}

b^{\log_e 2} = a^{\log_c c}

6a+

5bc

is equal to _______.

[10-Apr-2023 shift 1]

Your Answer:

Solution:

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

\ln a(\ln 2 + \ln a) = \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

x(\alpha + x) = y(y + z)

\alpha = \frac{xz}{y}

x\left(\frac{xz}{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

bc = 1 \text{ or } ab = 1

bc = 1 \text{ or } ab = 1

(a,b,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) } (a,b,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.

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