Solving $x-8 = x^{\log_{10}(2)}$

Question

I am trying to solve the equation $$x-8 = x^{\log_{10}(2)}$$

This is what I managed to do so far.

It is also pretty easy to find out that $x=10$ is a solution and that there are no more than 2 solutions (because $x^{\log_{10}(2)}$ is concave and $x-8$ is a line).

Answer 1

Here is how to "see" the only solution:

• $\lg 2 = \frac{\log_2 2}{\log_2 10} = \frac{1}{\log_2 10}$
• $x^{\lg 2} = 2^{\log_2 x}$

So, the equation becomes $$x-8 = x^{\lg(2)} \Leftrightarrow x - 2^{\frac{\log_2 x}{\log_2 10}} = 8$$

This equation has obviously the solution $x=10$.

The uniqueness follows (as already mentioned in other solution) from considering the monotonicity of $x - 2^{\frac{\log_2 x}{\log_2 10}}$.

Answer 2

$x=10$ is the solution since $$ 10^{\lg 2}=2=10-8 $$ To prove that there are no other solutions just plot left and right functions

Answer 3

$x =10$ is the only solution. Let $f(x) = x - 8 - x^{\lg2}, x\geq0$, then $f'(x)=1-\lg 2 x^{\lg 2 -1}$ and $f''(x) = (\lg2-(\lg 2)^2 )x^{\lg 2 -2}$ which is always non-negative for all $x\geq0$. Hence, $f'(x)$ is an increasing function.

Suppose $f(x)$ has more than $1$ root, we can denote each root by $x^*_i,\ i=1,2,\dots,n$. Note that $x^*_{i+1}>x^*_i>0$. Then since $f(x^*_i)=0$, we must have $x^*_i-8={x^*_i}^{\lg2}$ for each $i$. Then $f'(x^*_i)=1-\lg2+\lg2(\frac{8}{x^*_i})>0$. This means that at each root of $f(x)$, the derivative has to be postiive.

Now we know $f(0) = -8 <0$, and $f$ is continuous on $x\geq0$.

We also know $\lim_{x\rightarrow0} f'(x)=-\infty$ and $f'(x)$ is continuous on $x>0$, and $f'(x)$ is increasing.

Hence, $f'(x^*_1)>0$. For any $x>x^*_1$, $f'(x)\geq f'(x^*_1)>0$. If $f(x^*_2)=0$, by Rolle's thm, $f'(x)=0$ for some $x\in(x^*_1,x^*_2)$. but this is a contradiction.

Answer 4

Consider the function $$f(x)=x-x^k-8$$ where $k=\log_{10}(2) <1$ also implies that we work with $x \geq 0$.

Compute the derivatives $$f'x)=1-k\, x^{k-1} \qquad \text{and} \qquad f''(x)=-k(k-1)\,x^{k-2}$$

The first derivative cancels only once at $$x_*=\left(\frac{1}{k}\right)^{\frac{1}{k-1}}$$ and, since $k < 1$, by the second derivative test, this point corresponds to a minimum since $f''(x)> \,\, \forall k <1$. For $x >x_*$, $f'(x) >0$. Since, we also have $f(0)=-8$, there is only one root to the equation for any $k <1$.

I give you below the solution for various $k$

$$\left( \begin{array}{cc} k & \text{solution} \\ 0.05 & 9.11684 \\ 0.10 & 9.24914 \\ 0.15 & 9.39948 \\ 0.20 & 9.57106 \\ 0.25 & 9.76787 \\ 0.30 & 9.99496 \\ 0.35 & 10.2588 \\ 0.40 & 10.5680 \\ 0.45 & 10.9339 \\ 0.50 & 11.3723 \\ 0.55 & 11.9053 \\ 0.60 & 12.5658 \\ 0.65 & 13.4038 \\ 0.70 & 14.5010 \\ 0.75 & 16.0000 \\ 0.80 & 18.1769 \\ 0.85 & 21.6504 \\ 0.90 & 28.1834 \\ 0.95 & 45.9336 \end{array} \right)$$