Please help me in answering the following question
Find the number of real values of $x$ satisfying the equation:
$$\Large \left| 3 -x \right|^{ \log_7(x^2) - 7\log_x (49)} = (3-x)^3$$
I am able to get the answer as $1$ but the correct answer is $3$.
Kindly explain! :)
Since the LHS is non-negative, the RHS has to be non-negative, so $$3-x\ge0\iff x\le 3.$$
Case 1 : Obviously, $x=3$ is a solution.
Case 2 : For $3-x\gt 0\iff x\lt 3$, note that $|3-x|=3-x$.
If we let $t=\log_7 x$, then we have $$\begin{align}\log_7{x^2}-7\log_x{49}=3&\Rightarrow 2t-7\cdot\frac{2}{t}=3\\&\Rightarrow 2t^2-3t-14=0\\&\Rightarrow (2t-7)(t+2)=0\\&\Rightarrow t=\frac{7}{2},-2\\&\Rightarrow x=7^{\frac 72},7^{-2}.\end{align}$$ So, since $x\lt 3$, we have $x=7^{-2}=\frac {1}{49}$.
Hence, the answer is $$x=3,\frac{1}{49}.$$
The correct answer is actually wrong. $3$ is not "obviously" a solution. If you plug $3$ for $x$ into the base value it will give you $0$. But if you plug $3$ into the power/indix it will give you a negative number ($\approx-23.67$), and $0$ CANNOT be raised to the power of a negative integer. It is undefined.
But if you plug in $2$ for $x$, the base value will be $1$; and $1$ can be raised to any integer; and it will equal $1$ still. So $1$ is "obviously" a solution. But that is case 1.
Case 2, is $x<3$. So we just plug in for that and you'll get $7^{-2}$. (you can see the explaination for the second case from the answer provided by mathlove).
I don't know how to tackle the whole thing, but here is a start: $$\log_7{x^2}=2*\log_7 x$$ $$7\log_x 49=7\log_x 7^2=7*2\log_x 7=14\log_x 7$$ Now that we have those two pieces, notice the property of logarithms: $$\log_a x=\frac{\ln x}{\ln a}$$ We assign that to the second piece I wrote: $$14\log_x 7=14\frac{\log_7 7}{\log_7 x}=14\frac{1}{\log_7 x}$$ Combine them: $$2*\log_7 x -\frac{14}{\log_7 x}$$ Put them over a common denominator: $$\frac{2*(\log_7 x)^2-14}{\log_7 x}$$
If someone can take it from here, that would be appreciated. :)
