Solve equations of the form a^x = b using logarithms or other appropriate methods

Logarithmic and Exponential Functions – IGCSE 0606

Learning Objectives

By the end of this lesson you will be able to:

• state the definitions, domains, ranges and asymptotes of exponential and logarithmic functions,
• recognise that y = eˣ and y = ln x are exact inverses (reflection in the line y = x),
• sketch the basic shapes of y = aˣ and y = logₐ x and also the transformed forms y = k eⁿˣ + a and y = k logₐ(bx + c) + d,
• apply the laws of logarithms and the change‑of‑base formula,
• solve equations of the form aˣ = b (exact powers, using logarithms and using the change‑of‑base formula),
• solve simple logarithmic equations logₐ x = b,
• solve equations where the unknown occurs both inside and outside an exponential (e.g. 2ˣ = 3x) by iteration or graphically,
• identify cases with no real solution and express very large or very small results in scientific notation, and
• avoid the most common algebraic and calculator‑related mistakes.

Key Concepts

• Exponential function: \(y = a^{x}\) with \(a>0,\;aeq1\).
• Logarithmic function: \(x = \log_{a} b\) is the inverse of \(a^{x}=b\).
• Natural exponential and logarithm: \(y=e^{x}\) and \(y=\ln x\) (where \(e\approx2.71828\)). They are exact inverses; the graph of one is the reflection of the other in the line \(y=x\).
• Common logarithm: \(\log x = \log_{10}x\).

Properties of Exponential and Logarithmic Functions

Exponential	Logarithmic
\(a^{x+y}=a^{x}\,a^{y}\)	\(\log_{a}(xy)=\log_{a}x+\log_{a}y\)
\(a^{x-y}=a^{x}/a^{y}\)	\(\log_{a}\!\left(\dfrac{x}{y}\right)=\log_{a}x-\log_{a}y\)
\((a^{x})^{k}=a^{kx}\)	\(\log_{a}(x^{k})=k\log_{a}x\)
\(a^{0}=1,\;a^{1}=a\)	\(\log_{a}1=0,\;\log_{a}a=1\)

Change‑of‑Base Formula

For any positive bases \(aeq1\) and \(ceq1\):

\[ \log_{a}b=\frac{\log_{c}b}{\log_{c}a} \]

In practice the base \(c\) is usually 10 (common log) or \(e\) (natural log).

Graphs and Asymptotes

Basic shapes

• Exponential \(y=a^{x}\)
  • Domain: \(\mathbb{R}\); Range: \((0,\infty)\).
  • Horizontal asymptote: \(y=0\) (the x‑axis).
  • If \(a>1\) the curve rises to the right; if \(0
  • Intercepts: \((0,1)\); no x‑intercept.
• Logarithmic \(y=\log_{a}x\)
  • Domain: \((0,\infty)\); Range: \(\mathbb{R}\).
  • Vertical asymptote: \(x=0\) (the y‑axis).
  • If \(a>1\) the curve increases; if \(0
  • Intercepts: \((1,0)\); no y‑intercept.

Effect of coefficients, stretches and shifts

Multiplying or adding constants does not change the fundamental shape, but it moves or stretches the graph.

• Exponential with scale and shift: \(y = k\,e^{n x}+a\)
  • Vertical stretch/compression by \(|k|\) (if \(k<0\) the graph is reflected in the x‑axis).
  • Horizontal stretch/compression by \(\frac{1}{|n|}\) (if \(n<0\) the graph reflects in the y‑axis).
  • Vertical shift by \(a\) units.
• Logarithmic with scale and shift: \(y = k\,\log_{b}(c x+d)+a\)
  • Horizontal stretch/compression by \(\frac{1}{c}\) and shift left/right by \(-\frac{d}{c}\).
  • Vertical stretch/compression by \(|k|\) and shift up/down by \(a\).

Both transformed curves retain the same asymptotes (horizontal for exponentials, vertical for logarithms) unless a vertical shift removes the asymptote entirely.

Solving Equations of the Form \(\boldsymbol{a^{x}=b}\)

1. Check the domain: \(a>0,\;aeq1\) and \(b>0\). If \(b\le0\) there is no real solution.
2. Exact powers: If you can write \(b=a^{k}\) then \(x=k\) immediately (no calculator required).
3. Take logarithms of both sides. Choose the most convenient base:
   • Base‑\(a\) logarithm: \(\log_{a}(a^{x})=x\).
   • Common log \(\log\) or natural log \(\ln\) with the power rule \( \log(a^{x}) = x\log a\).
4. Isolate \(x\): \[ x=\frac{\log b}{\log a}\qquad\text{or}\qquad x=\frac{\ln b}{\ln a}. \]
5. Calculate keeping at least five significant figures until the final rounding. If the answer is very large or very small, express it in scientific notation (e.g. \(3.2\times10^{5}\)).

Advanced exponential equations (variable inside and outside)

When the unknown appears both as an exponent and elsewhere, an algebraic solution is usually not possible. Acceptable methods for IGCSE are:

• Graphical solution: Plot \(y=a^{x}\) and \(y= f(x)\) (the other side of the equation) on the same axes; the x‑coordinate of the intersection is the solution (to the required number of decimal places).
• Iteration / trial‑and‑error: Start with a reasonable guess, evaluate both sides, and refine the guess until the two sides agree to the required accuracy.

Example: solve \(2^{x}=3x\). By trial \(x\approx1.5\) gives \(2^{1.5}=2.828\) and \(3(1.5)=4.5\); increasing to \(x\approx1.8\) gives \(2^{1.8}=3.48\) and \(3(1.8)=5.4\). A graph shows the intersection at \(x\approx1.86\).

Solving Simple Logarithmic Equations \(\boldsymbol{\log_{a}x = b}\)

1. Rewrite using the definition of a logarithm: \(\log_{a}x = b \;\Longrightarrow\; a^{b}=x\).
2. Evaluate the power (exactly if possible, otherwise with a calculator).

Example: \(\log_{2}x = 5\) → \(x = 2^{5}=32\).

Common Mistakes to Avoid

• Assuming a solution exists when \(b\le0\); exponential functions never produce negative or zero values.
• Mis‑applying the power rule: the exponent becomes a *multiplier* (e.g. \(\log a^{x}=x\log a\)), not part of the argument.
• Mixing bases: \(\log_{a}beq\log_{b}a\) unless \(a=b\).
• Rounding intermediate results – keep at least five significant figures until the final answer.
• Forgetting that \(\log_{a}1=0\) and \(\log_{a}a=1\); these are useful checks.
• Neglecting the effect of coefficients and shifts when sketching transformed graphs.

Worked Examples

Example 1 – Exact power

Solve \(3^{x}=81\).

1. Write \(81=3^{4}\).
2. Since the bases are equal, the exponents are equal: \(x=4\).

Example 2 – Common logarithms

Solve \(3^{x}=20\) (no exact power).

1. Take common logs: \(\log(3^{x})=\log20\).
2. Apply the power rule: \(x\log3=\log20\).
3. Calculate \(\log20=1.3010,\;\log3=0.4771\).
4. \(x=\dfrac{1.3010}{0.4771}\approx2.73\) (two dp).

Example 3 – Natural logarithms

Solve \(e^{2x}=7\).

1. Take natural logs: \(\ln(e^{2x})=\ln7\).
2. Since \(\ln(e^{k})=k\), we have \(2x=\ln7\).
3. \(x=\dfrac{\ln7}{2}\approx\frac{1.9459}{2}=0.973\) (three dp).

Example 4 – Change‑of‑base

Solve \(2^{x}=5\) using natural logs.

\[ x=\frac{\ln5}{\ln2}\approx\frac{1.6094}{0.6931}=2.32\;(2\text{ dp}) \]

Example 5 – Graphical interpretation (negative exponent)

Find \(x\) such that \(4^{x}=0.5\).

• Using logs: \(x=\dfrac{\log0.5}{\log4}\approx\frac{-0.3010}{0.6021}=-0.50\).
• On a sketch of \(y=4^{x}\) the horizontal line \(y=0.5\) meets the curve at \(x\approx-0.5\).

Example 6 – Solving a logarithmic equation

Solve \(\log_{5}x = 3\).

• Rewrite: \(5^{3}=x\).
• Hence \(x=125\).

Example 7 – Equation with the variable both inside and outside an exponential

Solve \(2^{x}=3x\) (no algebraic solution).

• Graph \(y=2^{x}\) and \(y=3x\); the curves intersect at \(x\approx1.86\).
• Alternatively, use iteration:
  1. Start with \(x_{0}=1.5\) → \(2^{1.5}=2.828\), \(3x_{0}=4.5\) (right side too large).
  2. Increase to \(x_{1}=1.8\) → \(2^{1.8}=3.48\), \(3x_{1}=5.4\) (still too large).
  3. Try \(x_{2}=1.86\) → \(2^{1.86}=3.71\), \(3x_{2}=5.58\). The gap is narrowing; a calculator gives \(x\approx1.86\) to two dp.

Practice Questions

1. Solve \(4^{x}=64\) and give the exact answer.
2. Solve \(10^{x}=0.001\) using common logarithms.
3. Solve \(e^{2x}=7\) and give \(x\) correct to three decimal places.
4. Solve \(2^{x}=5\) using natural logarithms; round to two decimal places.
5. Show that \(3^{x}=27\) has the solution \(x=3\) without a calculator.
6. Using the change‑of‑base formula, evaluate \(\log_{2}12\) to three decimal places.
7. Sketch the graph of \(y=5^{x}\) and indicate its asymptote and intercepts.
8. Sketch the graph of \(y=2e^{3x}+1\); state the asymptote and the effect of the constants.
9. Solve \(\log_{3}x = 4\) and give the exact value of \(x\).
10. Solve \(2^{x}=3x\) by iteration or graphically; give the solution to two decimal places.

Answer Key (Teacher Use)

Q.	Solution
1	\(4^{x}=64\Rightarrow4^{x}=4^{3}\Rightarrow x=3\) (exact).
2	\(\log(10^{x})=\log0.001\Rightarrow x\log10=x=-3\).
3	\(e^{2x}=7\Rightarrow2x=\ln7\Rightarrow x=\dfrac{\ln7}{2}\approx0.973\) (3 dp).
4	\(2^{x}=5\Rightarrow x=\dfrac{\ln5}{\ln2}\approx2.32\) (2 dp).
5	\(3^{x}=27\Rightarrow27=3^{3}\Rightarrow x=3\) (no calculator).
6	\(\log_{2}12=\dfrac{\log12}{\log2}\approx\dfrac{1.0792}{0.3010}=3.585\) (3 dp).
7	Graph passes through \((0,1)\), is increasing, horizontal asymptote \(y=0\), y‑intercept \((0,1)\). Sketch to be supplied.
8	\(y=2e^{3x}+1\) has horizontal asymptote \(y=1\) (because \(2e^{3x}\to0\) as \(x\to-\infty\)). It is a vertically stretched (\(k=2\)) and horizontally compressed (\(n=3\)) version of \(e^{x}\), shifted up by 1.
9	\(\log_{3}x=4\Rightarrow x=3^{4}=81\) (exact).
10	Graphical/iterative solution gives \(x\approx1.86\) (2 dp).