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Logarithms test bite

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To complete this test bite we recommend you print off this question page and complete your working on paper. Then compare your answers and working to ours on the answer page.

Logarithms - test bite answers

Question

Evaluate:

1) \log _3 \;9

2) \log _5 \;125

3) \log _7 \;7

4) \log _8 \;1

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Answer

1) \log _3 \;9\; = \;2 since 3^2 \; = \;9

2) \log _5 \;125\; = \;3 since 5^3 \; = \;125

3) \log _7 \;7\; = \;1 since 7^1 \; = \;7

4) \log _8 \;1\; = \;0 since 8^0 \; = \;1

Question

Solve the following using the log button on your calculator. Give your answer correct to three decimal places.

1) 10^x \; = \;5

2) 10^{2x} \; = \;245

3) 10^{x - 1} \; = \;63

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Answer

1)

\eqalign{ 10^x \; & = & \;5 \Rightarrow x = \log _{10} 5 \cr\cr x& = & 0.699 \cr}

2)

\eqalign{ 10^{2x} \; & = & \;245 \Rightarrow 2x = \log _{10} 245 \cr\cr \Rightarrow x & = & {1 \over 2}\log _{10} 245 \cr\cr & = & 1.195 \cr}

3)

\eqalign{ 10^{x - 1} \; & = & \;63 \Rightarrow x - 1 = \log _{10} 63 \cr\cr \Rightarrow x & = & \log _{10} (63) + 1 \cr\cr & = & 2.799 \cr}

Question

Solve using the ln button on your calculator. Give your answer correct to three decimal places.

1) e^x \; = \;20

2) e^{3x} \; = \;9

3) e^{x + 4} \; = \;\sqrt {11}

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Answer

1)

\eqalign{ e^x \; & = & \;20 \cr\cr \Rightarrow x & = & \log _e 20 \cr\cr & = & 2.996 \cr}

2)

\eqalign{ e^{3x} \; & = & \;9 \cr\cr 3x & = & \log _e 9 \cr\cr x & = & {1 \over 3}\log _e 9 \cr\cr & = & 0.732 \cr}

3)

\eqalign{ e^{x + 4} \; & = & \;\sqrt {11} \cr\cr x + 4 & = & \log _e \sqrt {11} \cr\cr x & = & \log _e \left( {\sqrt {11} } \right) - 4 \cr\cr & = & - 2.801 \cr}

Question

Solve 4^x \; = \;32 giving your answer correct to one decimal place.

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Answer

Taking logs of both sides gives \eqalign{ \log _{10} 4^x & = & \log _{10} 32 \cr\cr x & = & {{\log _{10} 32} \over {\log _{10} 4}} \cr}

Use your calculator to find \log 32 \div \log 4 , giving x = 2.5 or \eqalign{ & (2^2 )^x = 32 \cr & 2^{2x} = 2^5 \cr & 2x = 5 \cr & x = 2.5 \cr}

Question

Simplify the following:

1) \log _a \;x\; + \;\log _a \;9

2) \log _a \;x^2 \; - \;\log _a \;x

3)

\log _a \;14\; + \;\log _a \;2x\; - \;\log _a \;7x

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Answer

1)

\eqalign{ && \log _a \;x\; + \;\log _a \;9 \cr && = \log _a 9x \cr}

2 )

There are two ways to simlpify this equation

\eqalign{ && \log _a \;x^2 \; - \;\log _a \;x \cr && = \log _a {{x^2 } \over x} \cr && = \log _a x \cr}

or

\eqalign{ && \log _a \;x^2 \; - \;\log _a \;x \cr && = 2\log _a x - \log _a x \cr && = \log _a x \cr}

3)

\eqalign{ && \log _a \;14\; + \;\log _a \;2x\; - \;\log _a \;7x \cr\cr && = \log _a {{14 \times 2x} \over {7x}} \cr\cr && = \log _a 4 \cr}

Question

Evaluate the following:

1) \log _8 \;16\; + \;\log _8 \;2

2) \log _3 \;81\; - \;\log _3 \;27

3) \log _5 \;100\; + \;\log _5 \;5\; - \;\log _5 \;2500

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Answer

1)

\eqalign{ && \log _8 \;16\; + \;\log _8 \;2 \cr\cr && = \log _8 \;16 \times 2 \cr\cr && = \log _8 \;32 \cr\cr && = {5 \over 3} \cr}

2 )

\eqalign{ && \log _3 \;81\; - \;\log _3 \;27 \cr\cr && = \log _3 \;{{81} \over {27}} \cr\cr && = \log _3 3 \cr\cr && = 1 \cr}

3)

\eqalign{ && \log _5 \;100\; + \;\log _5 \;5\; - \;\log _5 \;2500 \cr\cr && = \log _5 \;{{100 \times 5} \over {2500}} = \log _5 {1 \over 5} = \log _5 5^{ - 1} = - \log _5 5 \cr\cr && = - 1 \cr}

Question

If y\; = \;40\;e^{0.25t} find y when t = 10, giving your answer correct to three decimal places.

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Answer

Given y\; = \;40\;e^{0.25t} when t = 10, then y\; = \;40\;e^{0.25\times 10} = 40\;e^{2.5} = 487.300 to three decimal points

Question

If m_t \; = \;m_0 \;e^{kt} , find k when m_0 \; = \;60, m_t \; = \;20 and t = 5.

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Answer

From m_t \; = \;m_0 \;e^{kt} we can work out that 20 = 60e^{5t}

So e^{5t} = {{20} \over {60}} = {1 \over 3}

Take logs of base e of both sides, since the variable appears as a power of e .

\eqalign{ && \log _e e^{5t} = \log _e {1 \over 3} \cr\cr && 5t\log _e e = \log _e {1 \over 3} \cr\cr && 5t = \log _e {1 \over 3} \cr\cr && t = {1 \over 5}\log _e {1 \over 3} \cr}

k = -0.220 to three decimal places.

Question

If A\; = \;A_0 \;e^{ - kt}, find k if A\; = \;{1 \over 2}\;A_0 when t = 100.

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Answer

Putting A\; = \;{1 \over 2}\;A_0 into the equation A\; = \;A_0 \;e^{ - kt}, we get {{\rm{1}} \over {\rm{2}}}A_0 = A_0 e^{ - 100k}

Remove the common factor, giving e^{ - 100k} = {1 \over 2}

Take logs to base e of both sides.

\eqalign{ && \log _e e^{ - 100k} = \log _e {1 \over 2} \cr\cr && - 100k\log _e e = \log _e {1 \over 2} \cr\cr && - 100k = \log _e {1 \over 2} \cr\cr && k = - {1 \over {100}}\log _e {1 \over 2} \cr}

k = 0.007 to three decimal places.

Question

The equation of the line shown below has the form y\; = \;kx^n Determine the values of k and n.

Line intersecting y axis at 3 and gradient m

Line intersecting y axis at 3 and gradient m

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Answer

There are two methods you can use to solve this problem. We'll show you both.

  • Method 1

    Taking \log _{10} of both sides gives \eqalign{ && \log _{10} y = \log _{10} kx^n \cr\cr && = \log _{10} x^n + \log _{10} k \cr\cr && = n\log _{10} x + \log _{10} k \cr}

    Compare this with the equation of the line.

    \log _{10} y = 2\log _{10} x + 3

    We can see that n = 2

    \eqalign{ && \log _{10} k = 3 \cr && \Rightarrow k = 10^3 \cr && = 1000 \cr}

  • Method 2

    This method uses the equation y = mx + c

    From the diagram the equation is \eqalign{ && \log _{10} y = 2\log _{10} x + 3 \cr\cr && = \log _{10} x^2 + 3 \cr\cr && = \log _{10} x^2 + 3\log _{10} 10 \cr\cr && = \log _{10} x^2 + \log _{10} 10^3 \cr\cr && = \log _{10} 1000x^2 \cr} therefore y = 1000x^2

    and \eqalign{ && n = 2 \cr && k = 1000 \cr}

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