The curve y = ln x has:

x-intercept at (1, 0).

asymptote at x = 0.

• the + 3 raises the x-intercept of (1, 0) to
  (1, 3).
• the new x-intercept is at (e^-3, 0).
• the asymptote is unaffected.

The curve y = ln x has:

x-intercept at (1, 0).

asymptote at x = 0.

• the - 4 lowers the x-intercept of (1, 0) to
  (1, -4).
• the new x-intercept is at
  (e⁴ (= 54.6), 0).
• the asymptote is unaffected.

The curve y = ln x has:

x-intercept at (1, 0).

asymptote at x = 0.

• neither the x-intercept nor the asymptote is affected.
• at x = 2, the new point is (2, 4ln 2).

The curve y = ln x has:

x-intercept at (1, 0).

asymptote at x = 0.

• neither the x-intercept nor the asymptote is affected.
• the negative sign in the original equation shows the asymptote is at large positive values of y.
• at x = 1, the new point is (1, -4) because
  ln 1 = 0.

The curve y = 2ln 3x has:

x-intercept at (0.33, 0) as we put
3x = 1.

asymptote at x = 0.

• the x-intercept reduces to 0.33 (ln 3x = 0 gives 3x = 1) due to the curve increasing proportional to 3 not 1 while the asymptote is unaffected.
• at x = 2, the new point is (2, 2ln 6).

The square root simply reduces all the positive values and eliminates all negative values.

• x-intercept does not change from (1, 0).
• asymptote is removed because we can't take the square root of negative numbers..
• the curve shrinks due to all values being reduced to their square root.

The curve y = 3ln x - 4 has:

• x-intercept at (e^1.33, 0) as we put
  y = 0 so ln x = 4/3 = 1.33.
• asymptote is unchanged at x = 0.

The curve y = 3 + 4ln x has:

• x-intercept at (e^-0.75, 0) (about x = 0.47) as we put y = 0 so ln x = -3/4 = -0.75.
• asymptote is unchanged at x = 0.
• note how quickly the curve rises compared to the previous question.

The curve y = 2 - 2 ln x has:

• x-intercept at (e, 0) as we put
  y = 0 so ln x = 1.
• asymptote is unchanged at x = 0.
• curve comes down from the positive end of the y axis because of the negative in front of the log term.

The curve y = 5 + 4log_e x has:

• x-intercept at (e^-1.25, 0) as we put
  y = 0 so ln x = -5/4.
• asymptote is unchanged at x = 0.

The curve y = 2log_e (x + 4) has:

• x-intercept when x + 4 = 1 so at x = -3 .
• asymptote when x + 4 = 0 -
  so at x = -4.

The curve y = log_e (3x+2) has:

• x-intercept at (e, 0) as we put
  3x + 2 = 1 so x = -1/3 = -0.33.
• asymptote when 3x + 2 = 0
  so at x = -2/3 = -0.67 .
• y-intercept when x = 0 (y = ln 2).

The curve y = ln (5 - x) has:

• x-intercept when 5 - x = 1 so at x = 4.
• asymptote when 5 - x = 0 so at x = 5.
• the negative in front of the x swings the curve so its direction is now to the left.

The curve y = 2 ln (1 - 2x) has:

• x-intercept when 1 - 2x = 1 so at x = 0.
• asymptote when 1 - 2x = 0 so at x = 0.5.
• the negative in front of the x swings the curve so its direction is now to the left.

The curve y = 2ln (x + 1) + 3 has:

• x-intercept when ln (x + 1) = -1.5 -
  so when x = -1 + e^-1.5 = -0.78.
• asymptote when x + 1 = 0 so at x = -1.

The curve y = 3ln (1 - x) + 1 has:

• x-intercept when ln (1 - x) = -1/3 and then raised 1 - so when
  x = 1 - e^-0.33 = 0.28.
• asymptote when 1 - x = 0 so at x = 1.
• the negative in front of the x swings the curve so its direction is now to the left.

(ii) Solve the two equations simultaneously and show that your points of intersection are approximately reflected in your graph.

(i) See at right.

(ii) 2 ln x = ln x² = ln (5x - 6)

Equating the bits after the ln:

x2 - 5x + 6 = 0

(x - 2)(x - 3) = 0

x = 2 or x = 3

x = 2, y = ln 4 = 1.386

x = 3, y = ln 9 = 2.197

(ii) By drawing a second sketch of a relevant curve on your first set of axes, find the number of solutions for the equation ln x - x = -2.

(i) See at right.

(ii) Replace ln x with y so the equation now reads y - x = -2

So y = x - 2.

Draw this line (as shown) and count the number of points of intersection.
Answer: 2

So there are 2 solutions (and we are not asked even to approximate them) !!

25. is a difficult graph to draw.

• What two values of x can we see are not possible? x = 3 and x = 5.
• What about in-between 3 and 5 - say x = 4? Not possible as the log develops a negative sign for its argument.
• What about x < 3? As x gets closer to 3, the denominator becomes very small and so the value of the logarithm becomes very large. When x = 0, (5/3) > 1 so the log > 1.
  As x gets more negative, the denominator is always less than the numerator so the value of the log is always greater than 1.
• What about x > 5? The fraction has negative numerators and denominators but the denominator is always the larger value. So the value of the fraction is always positive but less than 1 - so the log is always negative.

The domain is therefore (-∞, 3) and (5, ∞).

As the value of the fraction approaches 1 when x approaches 3 or 5, the value of the log must approach 0.

The range is therefore .