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8. (a) Sketch the graph of $y = 7^x$, $x \\in \\mathbb{R}$, showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 3

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8.-(a)-Sketch-the-graph-of-$y-=-7^x$,-$x-\\in-\\mathbb{R}$,-showing-the-coordinates-of-any-points-at-which-the-graph-crosses-the-axes-Edexcel-A-Level Maths Pure-Question 9-2011-Paper 3.png

8. (a) Sketch the graph of $y = 7^x$, $x \\in \\mathbb{R}$, showing the coordinates of any points at which the graph crosses the axes. (b) Solve the equation $$7^{... show full transcript

Worked Solution & Example Answer:8. (a) Sketch the graph of $y = 7^x$, $x \\in \\mathbb{R}$, showing the coordinates of any points at which the graph crosses the axes - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 3

Step 1

Sketch the graph of $y = 7^x$

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Answer

To sketch the graph of the function y=7xy = 7^x, we first identify the points where the graph crosses the axes.

  • The graph crosses the y-axis at the point (0,70)=(0,1)(0, 7^0) = (0, 1).
  • The graph crosses the x-axis when y=0y = 0, but since an exponential function never reaches zero, it does not cross the x-axis.

The general shape of the graph is that it approaches the x-axis but never touches it, rising steeply as xx increases. A rough sketch shows an upward curve starting from the point (0,1)(0, 1).

Step 2

Solve the equation $7^{2x} - 4(7^x) + 3 = 0$

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Answer

To solve the equation, we can make a substitution. Let u=7xu = 7^x, then the equation becomes:

u24u+3=0u^2 - 4u + 3 = 0

This quadratic can be factored as:

(u3)(u1)=0(u - 3)(u - 1) = 0

Setting each factor to zero gives:

  1. u3=0u=37x=3x=log7(3)u - 3 = 0 \Rightarrow u = 3 \Rightarrow 7^x = 3 \Rightarrow x = \log_7(3)
  2. u1=0u=17x=1x=log7(1)=0u - 1 = 0 \Rightarrow u = 1 \Rightarrow 7^x = 1 \Rightarrow x = \log_7(1) = 0

Using the change of base formula, we find:

  • For x=log7(3)x = \log_7(3), we can express this in decimal form: x0.5646x \approx 0.5646: rounded to two decimal places is 0.560.56.

Thus, the solutions are:

  • x0.56x \approx 0.56
  • x=0x = 0

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