The temperature of a cooling liquid over time can be modelled by the exponential function
$$T(x)=60\left(\frac12\right)^\frac x{30}+20$$
where T(x) is the temperature, in degrees Celsius, and x is the elapsed time, in minutes.
Question I am trying to answer: Determine how long it takes for the temperature to reach 28 degrees Celsius.
I can't figure out how to isolate x. Thanks in advance for any help provided!
Apply $\ln(x^y) = y \ln(x)$: \begin{align} T(x) &= 60 \left(\frac{1}{2}\right)^{\large x/30} + 20 \iff \\ \ln\left(\frac{T - 20}{60}\right) &= \frac{x}{30} \ln \left( \frac{1}{2} \right) \iff \\ x(T) &= \cdots \end{align}
if you want to solve the equation $28=60\left(\frac{1}{2}\right)^{x/30}+20$
Simplify the above equation we get $\frac{8}{60}=\left(\frac{1}{2}\right)^{x/30}$
Taking the logarithm on both sides we obtain $\ln\left(\frac{2}{15}\right)=\frac{x}{30}\ln\left(\frac{1}{2}\right)$
Now i am sure you can isolate $x$ ?
