Logarithmic inequality: $7^{\sqrt{2x}}>20$

Question

We're working on logs and exponents in class, but I've never dealt with a situation where the equation isn't an equation but an inequality. Any help on how to solve this?! Thanks!

$$7^{\sqrt{2x}}>20$$

Answer 1

$$\sqrt{2x}\ln7>\ln{20}\Rightarrow \sqrt{2x}>\frac{\ln20}{\ln7}\Rightarrow x>(\frac{\ln20}{\ln7})^2/2$$

Don't seem to be much different from an equation.

Answer 2

$$ 7^{\sqrt{2x}} > 20. $$ If the base of the logarithmic function is bigger than $1$, rather than between $0$ and $1$, then the logarithmic function is increasing rather than decreasing. Consequently in the following step, "$>$" remains "$>$" rather than becoming "$<$": $$ \log \left(7^{\sqrt{2x}}\right) > \log 20. $$ $$ \sqrt{2x}\log 7 > \log 20. $$ If the base of the logarithmic function is bigger than $1$, rather than between $0$ and $1$, then the logarithm of $7$ is positive. It is for that reason that "$>$" remains "$>$" in the next step rather than changing to "$<$": $$ \sqrt{2x} >\frac{\log 20}{\log 7} = \log_7 20. $$ The next step works because the squaring function, applied to positive numbers, is increasing. If it were decreasing, then "$>$" would change to "$<$": $$ 2x>\left(\log_7 20\right)^2 $$ Finally $$ x>\frac 1 2 \left(\log_7 20\right)^2. $$ (And here we relied on the fact that $\dfrac12$ is positive.)

Answer 3

Logarithms to any base are functions that are strictly increasing, i.e. $$x>y\implies \log(x)>\log(y).$$ What this means in your case, is that you can "take the $\log$ of both sides", and preserve the inequality. So:

\begin{align*} 7^\sqrt{2x} &> 20 \\ \iff \log(7^\sqrt{2x}) &> \log(20) \\ \iff \sqrt{2x} \log(7) &> \log(20) \\ \iff \sqrt{2x} &> \frac{\log(20)}{\log(7)} \end{align*}

We want to square both sides, but we must be careful ($-3<1$, but $(-3)^2>1^2$). Thankfully, $\frac{\log(20)}{\log(7)}>0$, so both sides of the inequality are positive, so the inequality is preserved when we square.

\begin{align*} \iff 2x &> \left(\frac{\log(20)}{\log(7)}\right)^2 \\ \iff x &> \frac{1}{2}\left(\frac{\log(20)}{\log(7)}\right)^2 \end{align*}