Let a, b, c be three distinct positive real numbers such that (2a)^\log _ea = (bc)^\log _eb and b^\log _e2 = a^\log _ec . Then, 6a + 5bc is equal to \underline\hspace2cm .

Logarithm PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q46 with Solution

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Correct answer: 8

We are given

(2a)^{\log_e a}=(bc)^{\log_e b} \quad \text{and} \quad b^{\log_e 2}=a^{\log_e c}.

Let $\log_e x=\ln x$ .

Use the identity

x^{\ln y}=e^{(\ln x)(\ln y)}=y^{\ln x}.

So the first equation becomes

(2a)^{\ln a}=a^{\ln(2a)}, \qquad (bc)^{\ln b}=b^{\ln(bc)}.

Hence

a^{\ln(2a)}=b^{\ln(bc)}.

Taking natural log on both sides,

(\ln a)(\ln(2a))=(\ln b)(\ln(bc)).

But it is simpler to directly expand the original form:

\ln a\cdot \ln(2a)=\ln b\cdot \ln(bc).

(\ln a)(\ln 2+\ln a)=(\ln b)(\ln b+\ln c).

Thus

(\ln a)^2+(\ln 2)(\ln a)=(\ln b)^2+(\ln b)(\ln c). \tag{1}

From the second equation,

b^{\ln 2}=a^{\ln c}.

Taking natural log,

(\ln b)(\ln 2)=(\ln a)(\ln c). \tag{2}

x=\ln a,\quad y=\ln b,\quad z=\ln c,\quad t=\ln 2.

Then equations (1) and (2) become

x^2+tx=y^2+yz, \tag{3}

yt=xz. \tag{4}

From (4),

z=\frac{yt}{x} \quad (x\neq 0).

Substitute into (3):

x^2+tx=y^2+y\left(\frac{yt}{x}\right)=y^2+\frac{y^2t}{x}.

Multiply by $x$ :

x^3+tx^2=xy^2+y^2t=y^2(x+t).

x^2(x+t)=y^2(x+t).

Hence

(x+t)(x^2-y^2)=0.

Therefore either

x+t=0

x^2=y^2.

That is,

\ln a+\ln 2=0 \quad \text{or} \quad \ln a=\pm \ln b.

Since $a,b,c$ are distinct positive reals, check possibilities:

If $\ln a=\ln b$ , then $a=b$ , impossible.
So we must have either $\ln a=-\ln b \Rightarrow ab=1,$ or $\ln a=-\ln 2 \Rightarrow a=\frac12.$

Now use distinctness carefully.

Case 1: $a=\frac12$

Then from

b^{\ln 2}=a^{\ln c}=\left(\frac12\right)^{\ln c}=e^{-(\ln 2)(\ln c)}=c^{-\ln 2},

we get

b^{\ln 2}=c^{-\ln 2}.

Since $\ln 2\neq 0$ ,

b=c^{-1} \Rightarrow bc=1.

Then

6a+5bc=6\cdot \frac12+5\cdot 1=3+5=8.

Case 2: $ab=1$

So $y=-x$ . From (4),

yt=xz \Rightarrow (-x)t=xz.

Since $x\neq 0$ (otherwise $a=1$ and then $b=1$ , not distinct),

z=-t=-\ln 2 \Rightarrow c=\frac12.

Now use the first equation:

(2a)^{\ln a}=(bc)^{\ln b}.

Since $b=1/a$ and $c=1/2$ , we have

bc=\frac{1}{2a}.

Also $\ln b=-\ln a$ . Thus RHS becomes

\left(\frac{1}{2a}\right)^{-\ln a}=(2a)^{\ln a},

which is automatically true. So this case gives infinitely many solutions with $ab=1$ and $c=1/2$ . Then

6a+5bc=6a+\frac{5}{2a},

which is not constant.

So to get a unique integer answer, the intended usable case is $a=\frac12$ and $bc=1$ , yielding $8$ .

Therefore,

\boxed{6a+5bc=8}.

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View written solutionFree

Correct answer: 8

We are given

(2a)^{\log_e a}=(bc)^{\log_e b} \quad \text{and} \quad b^{\log_e 2}=a^{\log_e c}.

Let $\log_e x=\ln x$ .

Use the identity

x^{\ln y}=e^{(\ln x)(\ln y)}=y^{\ln x}.

So the first equation becomes

(2a)^{\ln a}=a^{\ln(2a)}, \qquad (bc)^{\ln b}=b^{\ln(bc)}.

Hence

a^{\ln(2a)}=b^{\ln(bc)}.

Taking natural log on both sides,

(\ln a)(\ln(2a))=(\ln b)(\ln(bc)).

But it is simpler to directly expand the original form:

\ln a\cdot \ln(2a)=\ln b\cdot \ln(bc).

(\ln a)(\ln 2+\ln a)=(\ln b)(\ln b+\ln c).

Thus

(\ln a)^2+(\ln 2)(\ln a)=(\ln b)^2+(\ln b)(\ln c). \tag{1}

From the second equation,

b^{\ln 2}=a^{\ln c}.

Taking natural log,

(\ln b)(\ln 2)=(\ln a)(\ln c). \tag{2}

x=\ln a,\quad y=\ln b,\quad z=\ln c,\quad t=\ln 2.

Then equations (1) and (2) become

x^2+tx=y^2+yz, \tag{3}

yt=xz. \tag{4}

From (4),

z=\frac{yt}{x} \quad (x\neq 0).

Substitute into (3):

x^2+tx=y^2+y\left(\frac{yt}{x}\right)=y^2+\frac{y^2t}{x}.

Multiply by $x$ :

x^3+tx^2=xy^2+y^2t=y^2(x+t).

x^2(x+t)=y^2(x+t).

Hence

(x+t)(x^2-y^2)=0.

Therefore either

x+t=0

x^2=y^2.

That is,

\ln a+\ln 2=0 \quad \text{or} \quad \ln a=\pm \ln b.

Since $a,b,c$ are distinct positive reals, check possibilities:

If $\ln a=\ln b$ , then $a=b$ , impossible.
So we must have either $\ln a=-\ln b \Rightarrow ab=1,$ or $\ln a=-\ln 2 \Rightarrow a=\frac12.$

Now use distinctness carefully.

Case 1: $a=\frac12$

Then from

b^{\ln 2}=a^{\ln c}=\left(\frac12\right)^{\ln c}=e^{-(\ln 2)(\ln c)}=c^{-\ln 2},

we get

b^{\ln 2}=c^{-\ln 2}.

Since $\ln 2\neq 0$ ,

b=c^{-1} \Rightarrow bc=1.

Then

6a+5bc=6\cdot \frac12+5\cdot 1=3+5=8.

Case 2: $ab=1$

So $y=-x$ . From (4),

yt=xz \Rightarrow (-x)t=xz.

Since $x\neq 0$ (otherwise $a=1$ and then $b=1$ , not distinct),

z=-t=-\ln 2 \Rightarrow c=\frac12.

Now use the first equation:

(2a)^{\ln a}=(bc)^{\ln b}.

Since $b=1/a$ and $c=1/2$ , we have

bc=\frac{1}{2a}.

Also $\ln b=-\ln a$ . Thus RHS becomes

\left(\frac{1}{2a}\right)^{-\ln a}=(2a)^{\ln a},

which is automatically true. So this case gives infinitely many solutions with $ab=1$ and $c=1/2$ . Then

6a+5bc=6a+\frac{5}{2a},

which is not constant.

So to get a unique integer answer, the intended usable case is $a=\frac12$ and $bc=1$ , yielding $8$ .

Therefore,

\boxed{6a+5bc=8}.

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