Solving logarithmic equations including x

Question

Let $$\log_3(x-2) = 6 - x$$ It's obvious drawing the graphs of the two functions that the only solution is $x=5$. But this is not really a proof, rather than observation.

How do you prove it algebraically?

Answer 1

We can solve this equation for $x$ in closed form as follows.

Write $\log_3 (x-2)=6-x$ as $\frac{\log (x-2)}{\log 3}=6-x$. Then we have

$$x-2=e^{\log 3[4-(x-2)]}=3^4e^{-(\log 3)\,(x-2)}\tag 1$$

Thus, multiplying $(1)$ by $\log 3$ and rearranging terms gives

$$(\log 3)(x-2)e^{(\log 3)(x-2)}=3^4\,\log 3$$

Recalling that Lambert's W function is defined as $z=W(z)e^{W(z)}$ gives

$$\begin{align} x&=2+\frac{W(3^4\,\log 3)}{\log 3}\\\\ &=2+3 \\\\ &=5 \end{align}$$

as was to be shown!

---

NOTE:

To show that $W(3^4\log 3)=3\log 3$ we use the interesting property of the W function that when $W(z)=x\log x$, $z=x^{x+1}\log x$. So, here we take $x=3$ and note that $W(3^4\log 3)=3\log 3$ ... as expected!

Answer 2

Finding the solution $x=5$ is just a matter of luck, try and error, approximation methods or just observation.

I guess that you are looking for a way to show that there are no more solutions. Just nothe that the function $$f(x)=\log_3(x-2)$$ is strictly increasing and the function $$g(x)=6-x$$ is strictly decreasing. Therefore

$$f(x)<1<g(x)$$ for $2<x<5$ and $$g(x)<1<f(x)$$ for $x>5$.

Answer 3

Since one has $x+\log_3(x-2)=6$, let $f(x)=x+\log_3(x-2)$. Note that $x\gt 2$.

Now, $$f'(x)=1+\frac{1}{(x-2)\ln 3}\gt 0.$$ Hence, since $y=f(x)$ is increasing for $x\gt 2$, we know that $f(x)=6$ has at most one real solution (you know that it is $x=5$).

Answer 4

Unfortunately there isn't a straightforward way to show this algebraically.

We know by inspection that $x=5$ is a solution since $$\log_3(5-2)=\log_3(3) = 1 = 6-5.$$

With a bit of calculus, you can demonstrate that these would not intersect again. This uses what is called the Rolle's Theorem.

Let $f(x) = \log_3(x-2) - 6 + x$. If $y$ satifies your equation, then $f(y)=0$.

We know that $f(5)=0$ for example.

Rolle's Theorem says that if there is another solution $y$, then there is a point $\xi$ in the interval $[5,y]$ (we are assuming $y>5$ here, but the same argument works for $y<5$) for which $f'(\xi)=0$.

However $f'(x) = \frac{1}{(x-2)\ln(3)} + 1 = \frac{1+(x-2)\ln(3)}{(x-2)\ln(3)}$ which means $1+(\xi-2)\ln(3) =0$ or $\xi =2-\ln(3)^{-1}$. This is outside the domain of $\log_3(x-2)$, which is a contradiction.

Thus the only solution is $x=5$.

Answer 5

You don't: other than in some special cases equations involving logarithms, exponentials, or trigonometric functions (to name a few) don't have algebraic solutions. You can prove it analytically by showing that the LHS is strictly increasing with a zero at $x = 3$, while the RHS is strictly decreasing and positive at $x = 3$.