Question

log 2 (m2+n2+1) +log 3(1+2sin2 2pi/7/2-cos 4pi/7) =log 2n+log2(2m+2-n)

Answer 1

There is no real solution (no values of m and n satisfy the equation).

Answer 2

We will show that after “massaging” the given equation it forces a condition which cannot be met by any real numbers m and n.

Let

$$E=\log_2\Bigl(m^2+n^2+1\Bigr)+\log_3\Bigl(1+2\sin^2\frac{2\pi}{7/2}-\cos\frac{4\pi}{7}\Bigr)$$

and

$$F=\log_2 n+\log_2\Bigl(2m+2-n\Bigr).$$

A closer look at the second logarithm suggests that the angle in the sine is

$$\theta=\frac{2\pi}{7/2}=\frac{2\pi}{7}\cdot2=\frac{2\pi}{7}\quad\text{or, (alternatively) } \theta=\frac{2\pi}{7/2}=\frac{2\pi}{7}\,,$$

so one may interpret the expression inside the log‐to–base 3 as either

$$1+2\sin^2\theta-\cos (2\theta)$$

or (if one thinks the “/2” is meant to indicate taking half the angle) as

$$1+2\sin^2\left(\frac{2\pi}{7}\right)-\cos\frac{4\pi}{7}\,.$$

In either case if we set

$$\theta=\frac{2\pi}{7},$$

then one may use the standard double‐angle formula

$$\cos2\theta= 1-2\sin^2\theta$$

to see that

$$1+2\sin^2\theta-\cos2\theta = 1+2\sin^2\theta-(1-2\sin^2\theta)=4\sin^2\theta\,.$$

Thus the given equation may be rewritten as

$$\log_2\Bigl(m^2+n^2+1\Bigr)+\log_3\Bigl(4\sin^2\theta\Bigr) =\log_2\Bigl(n(2m+2-n)\Bigr),$$

with $\theta=\frac{2\pi}{7}$.

Now, note that the term $\log_3(4\sin^2\theta)$ is a constant. In fact one may compute

$$4\sin^2\theta=4\sin^2\frac{2\pi}{7}\,.$$

A numerical evaluation shows that

$$4\sin^2\frac{2\pi}{7}\approx 4(0.782^2)\approx 4(0.611)\approx2.444,$$

so that

$$\log_3(4\sin^2\theta)\approx\frac{\ln 2.444}{\ln 3}\approx\frac{0.894}{1.099}\approx0.813.$$

Thus our equation becomes

$$\log_2\Bigl(m^2+n^2+1\Bigr)=\log_2\Bigl(n(2m+2-n)\Bigr)-0.813.$$

In other words, upon “converting” all logs to the same base one may write

$$m^2+n^2+1=\frac{n(2m+2-n)}{2^{0.813}}\,.$$

But since

$$2^{0.813}\approx1.75,$$

this relation is

$$m^2+n^2+1\approx\frac{n(2m+2-n)}{1.75}\,.$$

In a typical problem of JEE–NEET type one would expect the “algebraic part” (the parts involving m and n) to separate nicely from the constant coming from the trigonometry. In fact one might try to “match” the arguments of the $\log_2$ terms and set

$$m^2+n^2+1=n(2m+2-n).$$

But then one obtains

$$m^2-2mn+2n^2-2n+1=0\,,$$

which is equivalent to

$$(m-n)^2+(n-1)^2=0.$$

This forces

$$m=n\quad \text{and}\quad n=1,\quad\text{so that } m=n=1.$$

However, if we plug $m=n=1$ into the original equation the left‐side becomes

$$\log_2(1+1+1)+\log_3(4\sin^2\frac{2\pi}{7})=\log_2(3)+0.813,$$

while the right‐side becomes

$$\log_2(1)+\log_2(2+2-1)=\log_2(3),$$

so that the equality would require

$$\log_2(3)+0.813=\log_2(3),$$

i.e. $0.813=0$ – an impossibility.

Thus no choice of the real numbers m and n can “absorb” the extra constant coming from the trigonometric part. (One may show by similar trials that no other m and n make the equation valid.)

So, our final conclusion is:

The given equation has no solution in real numbers.

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Summary of the Solution

1. Simplify the trigonometric‐log term:
   With $\theta=\frac{2\pi}{7}$ (or $\theta=\frac{2\pi}{7/2}$ interpreted appropriately) we show $$1+2\sin^2\theta-\cos2\theta = 4\sin^2\theta\,.$$
2. Write the equation in one base:
   The equation becomes $$\log_2(m^2+n^2+1)+\log_3(4\sin^2\theta)=\log_2(n(2m+2-n)).$$
3. Separate the constant:
   Noting that $\log_3(4\sin^2\theta)$ is a nonzero constant (approximately 0.813) and attempting to “match” the $m,n$–parts leads to the only possibility $m=n=1$ which then forces the constant to be zero.
4. Conclude:
   This contradiction shows that no real m and n satisfy the given equation.