Water cooling in fridge

I'm suppose to solve a problem that goes like this.

The graph for the following function f given by $f(x) = 115.82 \cdot 0.94^x + 5$, with $x \geq 5$, gives the temperature of the water after it's been placed in the fridge. Find by calculation what time period the temperature is above 60. Now for me, it appears that the solution is the inequality shown in the picture above.

$$115.82 \cdot 0.94^x + 5 > 60$$

If I solve this I get
$$x > 12.04$$

It should be less than that. Not greater. Does taking the log on both sides require one to flip the thingy thing?

$$\begin{aligned} 115.82 \cdot 0.94^x + 5 &> 60 \\ \frac{115.82 \cdot 0.94^x}{115.82} &> \frac{55}{115.82} \\ 0.94^x &> \frac{55}{115.82} \\ \log{0.94^x} &> \log{\frac{55}{115.82}} \\ x \cdot \log{0.94} &> \log{\frac{55}{115.82}} \\ x &> \frac{\log \frac{55}{115.82}}{\log 0.94} \\ \end{aligned}$$

  • 6
    $\begingroup$ Did you remember that $\ln(0.94)$ is negative, and so when you divide by it to isolate $x$ you need to change the inequality sign? $\endgroup$ – Arturo Magidin Nov 21 '11 at 21:25
  • $\begingroup$ By the way, the "thingy" is the "inequality sign", or the "greater-than-or-equal sign". $\endgroup$ – Arturo Magidin Nov 21 '11 at 22:11

From $$115.82 \times 0.94^x + 5\gt 60$$ we get (switching to $\log$ base 10): $$\begin{align*} (115.82)(0.94^x) &\gt 55\\ 0.94^x &\gt \frac{55}{115.82}\\ \log(0.94^x) &\gt \log\frac{55}{115.82}\\ x\log(0.94) &\gt \log (55) - \log(115.82). \end{align*}$$ Now, since $0.94\lt 1$, then $\log(0.94)\lt \log(1) = 0$; so if we divide by $\log(0.94)$, we are dividing by a negative number, which reverses the inequality. We get: $$x \lt \frac{\log(55)-\log(115.82)}{\log(0.94)}\approx 12.0355.$$

Added. The same thing happens with any logarithm with base $a\gt 1$ (e.g., with $\ln$, the natural logarithm).

For logarithms with base $a$, $0\lt a\lt 1$, taking logarithms on both sides of an inequality $0\lt r\lt s$ reverses the inequality, but logarithms of numbers smaller than $1$ are positive and those of numbers larger than $1$ are negative.

  • $\begingroup$ In this course "ln" is not used, but is it the same? If i do ln 0.94 and log 0.94 the numbers differ, but for this purpose can it be ignored? $\endgroup$ – Algific Nov 21 '11 at 21:41
  • 2
    $\begingroup$ In this calculation you will get the same answer no matter what kind of logarithm you use. $\log_ar/\log_as$ takes on the same value, no matter what base $a$ is used. $\endgroup$ – Gerry Myerson Nov 21 '11 at 21:52
  • $\begingroup$ @Algific: For any base $a\gt 1$, $\log_a(r)\lt 0$ if $0\lt r\lt 1$, and $\log_a(r)\gt 0$ if $r\gt 1$. For logarithm bases with $0\lt a\lt 1$, you have $\log_a(r)\gt 0$ if $0\lt r\lt 1$, and $\log_a(r)\lt 0$ if $1\lt r$. I assume you will be using the common logarithm (logarithm base 10); the inequalities are the same: $\log(r) \lt 0$ if $0\lt r\lt 1$, and $\log(r)\gt 0$ if $r\gt 1$. $\endgroup$ – Arturo Magidin Nov 21 '11 at 21:56

"Does taking the log on both sides require one to flip the thingy thing?"

No, but multiplying (or dividing) by a negative number does.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.