AI HL · 12.6
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Mixed exponent and logarithm practice
12.6.3
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Given that \(x=\log_{10}a\), \(y=\log_{10}b\) and \(z=\log_{10}c\), write \(\log_{10}\!\left(\dfrac{a^{2}\sqrt{b}}{c^{3}}\right)\) in terms of x, y and z.
Have a go first. Then open the working one line at a time.
\(\log_{10}a^{2}+\log_{10}\sqrt{b}-\log_{10}c^{3}\)
Multiplying adds the logs. Dividing subtracts.
\(\log_{10}a^{2}+\log_{10}b^{1/2}-\log_{10}c^{3}\)
\(\sqrt{b}=b^{1/2}\).
\(2\log_{10}a+\tfrac12\log_{10}b-3\log_{10}c\)
\(\log_{10}a^m=m\log_{10}a\).
\(2x+\tfrac12y-3z\)
Replace each log with its letter.
Solve \(\log_{10}(2x+1)=2\).
Have a go first. Then open the working one line at a time.
\(2x+1=10^{2}\)
\(\log_{10}A=2\) means \(A=10^2\).
\(2x=99\), so \(x=49.5\)
Subtract 1, then divide by 2.
\(\log_{10}(2\times49.5+1)=\log_{10}100=2\)
The check puts x back in.
Solve \(3\times2^{x}+5=40\). Give x to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(3\times2^{x}=35\), so \(2^{x}=\dfrac{35}{3}\)
Subtract 5 first, then divide by 3.
\(x\ln2=\ln\dfrac{35}{3}\)
\(\ln 2^x=x\ln2\). Any base of log works; ln is on the calculator.
From the calculator: \(x=\dfrac{\ln(35/3)}{\ln2}=3.54\)
Give 3 significant figures.
Solve \(3^{x+2}=2^{3x-1}\), giving x in the form \(\dfrac{\ln p}{\ln q}\). Then give x to 3 significant figures.
Have a go first. Then open the working one line at a time.
\((x+2)\ln3=(3x-1)\ln2\)
Bracket the whole exponent.
\(x\ln3+2\ln3=3x\ln2-\ln2\)
Multiply out both brackets.
\(2\ln3+\ln2=x(3\ln2-\ln3)\)
Put the x terms on one side.
\(x=\dfrac{\ln18}{\ln(8/3)}\)
\(2\ln3+\ln2=\ln18\) and \(3\ln2-\ln3=\ln\frac83\).
From the calculator: \(x=2.95\)
The number of followers of a channel is modelled by \(F=1200\times1.08^{t}\), where t is the time in months. Use the calculator to find when F reaches 3000.
Have a go first. Then open the working one line at a time.
\(1200\times1.08^{t}=3000\)
Put F = 3000.
Solve for t: \(t=11.9\)
Use the calculator's numerical solver. Give 3 significant figures.
\(1200\times1.08^{11.9}\approx3000\)
State the answer with units: months.
The mass of a radioactive sample is modelled by \(M=80e^{-0.035t}\) mg, where t is the time in days. Find the half-life of the sample.
Have a go first. Then open the working one line at a time.
\(80e^{-0.035t}=40\)
The start mass is 80 mg, so half is 40 mg.
\(e^{-0.035t}=0.5\)
Divide by 80. The start mass cancels, so the half-life does not depend on it.
\(-0.035t=\ln0.5=-\ln2\)
\(\ln0.5=-\ln2\).
\(t=\dfrac{\ln2}{0.035}=19.8\) days
Give units. After about 19.8 days the mass is halved.
A culture of cells grows according to \(N=N_0e^{kt}\), where t is the time in hours. There are 500 cells at \(t=0\) and 1850 cells at \(t=6\). Find k. Then find when there are 10 000 cells.
Have a go first. Then open the working one line at a time.
\(N_0=500\)
At t = 0, \(e^{0}=1\).
\(500e^{6k}=1850\Rightarrow e^{6k}=3.7\)
Divide by 500.
\(6k=\ln3.7\), so \(k=0.218\)
Keep the full value of k on the calculator.
\(e^{kt}=20\Rightarrow t=\dfrac{\ln20}{k}=13.7\) hours
Divide 10 000 by 500, then take ln.
Given that \(7\times6^{x}=35\times3^{x}\): (a) write down the exact value of \(2^x\); (b) hence find x.
Have a go first. Then open the working one line at a time.
\(6^{x}=2^{x}\times3^{x}\)
\(6=2\times3\), so \(6^x=2^x3^x\).
\(7\times2^{x}\times3^{x}=35\times3^{x}\), so \(2^{x}=5\)
Divide both sides by \(7\times3^x\).
\(x=\dfrac{\ln5}{\ln2}\)
From the calculator: \(x=2.32\)
Solve \(\log_{x}16=\dfrac43\).
Have a go first. Then open the working one line at a time.
\(x^{4/3}=16\)
\(\log_xk=n\) means \(x^n=k\).
\(x=16^{3/4}\)
Raise both sides to the reciprocal of 4/3.
\(x=(2^{4})^{3/4}=2^{3}=8\)
Write 16 as a power of 2 first.