The logarithmic function：練習題

Solve the equation \(10^{2x} = 450\) for \(x\). Give your answer correct to 3 significant figures.

Solve the equation \(5 \cdot 3^{2x} = 80\). Give your answer correct to 3 significant figures.

If \(10^y = x + 3\), which of the following correctly expresses \(y\) in terms of \(x\)?

Solve the logarithmic equation \(\log_{10}(x) + \log_{10}(x - 3) = 1\).

Solve the logarithmic equation \(\log_{10}(x - 4) + \log_{10}(x + 5) = 1\) for \(x\).

Solve the equation $$4^x = 25$$ for $$x$$, giving your answer correct to 3 significant figures.

Solve the equation $$\log_{10}(x+2) + \log_{10}(x-1) = 1$$.

A sum of $$5000 is invested at an annual interest rate of $$4% compounded annually. Find the number of years it will take for the investment to reach $$8000. Give your answer correct to one decimal place.

A sum of 3000 is invested in an account that pays 4.2% interest per annum, compounded annually. The value of the investment, \(V\), after \(n\) years is given by the formula \(V = 3000(1.042)^n\).

(a) Calculate the value of the investment after 6 years, giving your answer correct to 2 decimal places.
(b) Use logarithms to calculate the number of years it will take for the value of the investment to reach 5000. Give your answer correct to 3 significant figures.
(c) Solve the equation \(5^{2x-1} = 8^x\) for \(x\), giving your answer correct to 3 significant figures.

The population of a certain bacteria, $$P$$, at time $$t$$ hours, is modelled by the formula $$P = P_0 e^{kt}$$, where $$P_0$$ is the initial population and $$k$$ is the growth constant.

(a) Initially, there are 500 bacteria. After 3 hours, the population has grown to 1200 bacteria. Show that the growth constant $$k$$ is approximately $$0.292$$, correct to 3 significant figures.

(b) Using the value of $$k = \frac{\ln(2.4)}{3}$$ (exact form), find the time, in hours, it takes for the population to reach 5000 bacteria. Give your answer correct to 1 decimal place.

(c) A third type of bacteria has its population, $$R$$, modelled by the formula $$R = 100 \cdot 2^{0.5t}$$. Find the time, in hours, when the population of the first type of bacteria ($$P$$) is equal to the population of this third type of bacteria ($$R$$). Give your answer correct to 1 decimal place.