Solutions of $2^x 7^{1/x}\le 14$

Question

The solution is supposed to be $(-\infty,0)$ and $[1,\log_2 7]$.

What I get when solving the problem is $(-\infty, \log_2 7]$.

Where did I get it wrong? I start by dividing both sides by 14, then get $2$ and $7$ on opposite sides, where I take raise both sides to $\frac1{x-1}$ (Am I allowed to do that?). This leads me to an expression that when logarithmed gives me the mentioned solution. Where did I go wrong?

Answer 1

If $x<1$ the function $f(t)=t^{1/(x-1)}$ is decreasing for $t>0$. This means that if you raise both sides of inequality to $\frac1{x-1}$, its direction is reversed.

If I have understood you correctly, you have arrived at $$2^{x-1}\le 7^{(x-1)/x}$$

Now, you should take logarithms in base $2$ (which is an increasing function) to get $$x-1\le\frac{x-1}x\log_27$$

or $$\frac{-x^2+x(1+\log_27)-\log_27}{x}\ge 0$$

Answer 2

Presumably, this is your argument:

$$2^{x-1}7^{\frac{1}{x}-1} \leq 1$$ $$2^{x-1}\leq 7^{1-\frac{1}{x}}$$

So:

$$2^{x-1}\leq \left(7^{1/x}\right)^{x-1}\tag{1}$$

Now there should be two cases, one where $x-1<0$ and one where $x-1\geq 0$.

If $x-1<0$, then $2\geq 7^{1/x}$, or $\log_7{2}\leq \frac{1}x$. Again there are two cases:

a. If $x< 0$, you need $x\leq \log_2 7$, which is always true.

b. If $0< x<1$ then $x\geq \log_2 7$, which is never true, since $\log_2 7>1$.

(Note, $x=0$ is not of concern.)

Finally, you have to consider the case when $x-1\geq 0$. Then from (1):

$$2\leq 7^{1/x}$$ or $2^x\leq 7$ or $x\leq \log_2 7$.

Again, without knowing what you did exactly, I don't know which case you missed, but it seems likely that you forgot that, in the case when $x-1<0$:

$$\log_7 2\leq \frac{1}{x}$$ has two cases, whether $x$ is negative or not.