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JEE Mathematics 2023 Question with Solution

Let aa, bb, cc be three distinct positive real numbers such that (2a)log⁡ea=(bc)log⁡eb\left(2a\right)^{\log_e a} = \left(bc\right)^{\log_e b} and blog⁡2(ea)=alog⁡e(c)b\log_2\left(e^a\right) = a\log_e\left(c\right). Then 6a+5bc6a + 5bc is equal to:

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given:

(2a)log⁡ea=(bc)log⁡eb\left(2a\right)^{\log_e a} = \left(bc\right)^{\log_e b}

and

blog⁡2(ea)=alog⁡e(c)b\log_2\left(e^a\right) = a\log_e\left(c\right)

Find: 6a+5bc6a + 5bc

Let

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

From

(2a)log⁡ea=(bc)log⁡eb\left(2a\right)^{\log_e a} = \left(bc\right)^{\log_e b}

taking natural logarithm gives

x(α+x)=y(y+z)x\left(\alpha + x\right) = y\left(y + z\right)

Also, from

blog⁡2(ea)=alog⁡e(c)b\log_2\left(e^a\right) = a\log_e\left(c\right)

the solution uses

αy=xz\alpha y = xz

Substituting

α=xzy\alpha = \frac{xz}{y}

into

x(α+x)=y(y+z)x\left(\alpha + x\right) = y\left(y + z\right)

we get

x2(z+y)=y2(y+z)x^2\left(z + y\right) = y^2\left(y + z\right)

so

(y+z)(x2−y2)=0\left(y + z\right)\left(x^2 - y^2\right) = 0

Hence either

y+z=0y + z = 0

or

x=−yx = -y

which correspond respectively to

bc=1bc = 1

and

ab=1ab = 1

For the valid distinct positive real values obtained in the solution, the expression evaluates to

6a+5bc=3+5=86a + 5bc = 3 + 5 = 8

Therefore, the required numerical value is 88.

Answer from extracted the solution

The solution concludes with Correct Answer: 88 and ends with

8\boxed{8}

Therefore, the required answer is 88.

Common mistakes

  • A common mistake is to treat log⁡2(ea)\log_2\left(e^a\right) as aa. This is incorrect because the base is 22, so log⁡2(ea)=alog⁡2e\log_2\left(e^a\right) = a\log_2 e. Always account for the change of base factor.

  • Another mistake is to conclude directly from x2=y2x^2 = y^2 that x=yx = y only. This is incomplete because x2=y2x^2 = y^2 gives both x=yx = y and x=−yx = -y. Here the relevant branch used in the solution is x=−yx = -y.

  • Students may confuse y+z=0y + z = 0 with b+c=0b + c = 0. Since y=ln⁡by = \ln b and z=ln⁡cz = \ln c, the correct interpretation is ln⁡b+ln⁡c=0\ln b + \ln c = 0, hence ln⁡(bc)=0\ln(bc) = 0 and therefore bc=1bc = 1.

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