Basics & LogarithmshardFree

Greatest value of ln x ln z given ln x + log_y z = 3 and ln y + log_x z = 4 - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
The greatest value of ln⁡(x)ln⁡(z)\ln\left(x\right)\ln\left(z\right) provided that ln⁡(x)+log⁡y(z)=3\ln\left(x\right)+\log_y\left(z\right) = 3 and ln⁡(y)+log⁡x(z)=4\ln\left(y\right)+\log_x\left(z\right) = 4 where x,y,z>1x,y,z>1.
Solution
Answer: 5.33 (± 0.01)
Step 1: Let ln⁡x=a\ln x = a, ln⁡y=b\ln y = b, ln⁡z=c\ln z = c. Given x,y,z>1x, y, z > 1, so a,b,c>0a, b, c > 0.
log⁡yz=cb,log⁡xz=ca\log_y z = \frac{c}{b}, \qquad \log_x z = \frac{c}{a}
∴ a+cb=3  ...(1),b+ca=4  ...(2)\therefore\ a + \frac{c}{b} = 3 \ \ ...(1), \qquad b + \frac{c}{a} = 4 \ \ ...(2)
 The required value is the greatest value of acac. Step 2: From (1), cb=3−a\dfrac{c}{b} = 3 - a. Here b,c>0b, c > 0, so 3−a>03 - a > 0 and
b=c3−a,0<a<3b = \frac{c}{3-a}, \qquad 0 < a < 3
 Step 3: Substituting in (2),
c3−a+ca=4  ⟹  c⋅a+(3−a)a(3−a)=4  ⟹  c=43 a(3−a)\frac{c}{3-a} + \frac{c}{a} = 4 \implies c \cdot \frac{a + (3-a)}{a(3-a)} = 4 \implies c = \frac{4}{3}\,a(3-a)
∴ ac=43 a2(3−a)\therefore\ ac = \frac{4}{3}\,a^2(3-a)
 Step 4: a2, a2, 3−a\dfrac{a}{2},\ \dfrac{a}{2},\ 3-a are positive and their sum is 33, so by AM-GM
a2⋅a2⋅(3−a)≤(33)3=1  ⟹  a2(3−a)≤4\frac{a}{2}\cdot\frac{a}{2}\cdot(3-a) \le \left(\frac{3}{3}\right)^3 = 1 \implies a^2(3-a) \le 4
 Step 5:
ac=43 a2(3−a)≤43×4=163ac = \frac{4}{3}\,a^2(3-a) \le \frac{4}{3} \times 4 = \frac{16}{3}
 Equality holds when a2=3−a\dfrac{a}{2} = 3 - a, i.e. a=2a = 2. Step 6: a=2  ⟹  c=43(2)(1)=83a = 2 \implies c = \dfrac{4}{3}(2)(1) = \dfrac{8}{3} and b=c3−a=83b = \dfrac{c}{3-a} = \dfrac{8}{3}, all positive. Check in (2): b+ca=83+43=4b + \dfrac{c}{a} = \dfrac{8}{3} + \dfrac{4}{3} = 4. So the bound is attained at x=e2, y=z=e8/3x = e^2,\ y = z = e^{8/3}, and
ln⁡x⋅ln⁡z=2×83=163\ln x \cdot \ln z = 2 \times \frac{8}{3} = \frac{16}{3}
 Answer: 163\dfrac{16}{3}.
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