IMPLICIT DIFFERENTIATION

how to solve equations involving logarithms

Solving equations involving logarithms generally require the application of the properties of logarithms. In most cases the use of the property – \(\log_a⁡x=\log_a⁡y→x=y\) is needed to obtain an equation (linear, quadratic or otherwise) which can then be easily solved.

Solving equations involving logarithms generally require the application of the properties of logarithms. In most cases the use of the property – \(\log_a⁡x=\log_a⁡y→x=y\) is needed to obtain an equation (linear, quadratic or otherwise) which can then be easily solved.

LESSON 1

Solve the equation \(3 \lg⁡(x-1)=\lg⁡27\)

SOLUTION

\(3 \lg⁡(x-1)=\lg⁡27\)

• Use the property \(\log_a⁡P^b =b \log_a⁡P\)

\(\lg⁡(x-1)^3=\lg⁡3^3\)

• Use the fact that \(\log_a⁡x=\log_a⁡y⟹x=y\)

\((x-1)^3=3^3\)

• Take the cube root of both sides.

\(x-1=3\)

\(x=4\)

---

LESSON 2

Find the value(s) of \(x∈R\) which satisfy \(2 \log_3⁡x=\log_3⁡(x+12) \)

SOLUTION

\(2 \log_3⁡x=\log_3⁡(x+12)\)

• Use the property \(\log_a⁡P^b=b \log_a⁡P\)

\(\log_3⁡x^2 =\log_3⁡(x+12)\)

• Use the fact that \(\log_a⁡x=\log_a⁡y⟹x=y\)

\(x^2=x+12 \)

\(x^2-x-12=0\)

\((x-4)(x+3)=0 \)

\(x=-3,4\)

• Ensure that answers are valid.

\(x=-3\) is INVALID since substituting this value into the original equation will result in us having to find logarithms of a negative number.

---

LESSON 3

Solve the equation \(\log_3⁡(4x)+\log_3⁡(x-1)=1\)

SOLUTION

We need to write one logarithmic expression on the left and one logarithmic expression on the right so that we can equate.

\(\log_3⁡(4x)+\log_3⁡(x-1)=1\)

• Use the properties \(\log_a⁡P+\log_a⁡Q=\log_a⁡PQ\) and \(\log_a⁡a\).

\(\log_3⁡[(4x)(x-1)]=\log_3⁡3\)

• Use the fact that \(\log_a⁡x=\log_a⁡y⟹x=y\)

\(4x(x-1)=3\)

\(4x^2-4x=3\)

\(4x^2-4x-3=0\)

\((2x-3)(2x+1)=0\)

\(x={3 \over 2}\),\(-1 \over 2\)

• Ensure the validity of the answers.

\(x=-{1 \over 2}\) is INVALID since substituting this value into the original equation will result in us having to find logarithms of a negative number.

---

LESSON 4

Solve the equation \(\lg⁡(10x)-\lg⁡(x-9)=2\)

SOLUTION

\(\lg⁡(10x)-\lg⁡(x-9)=2\)

• Since we need to have logarithms on both sides of the equation we need to rewrite 2 using \lg (base 10)

\(\lg⁡(10x)-\lg⁡(x-9)=2 \lg⁡10\)

• Use the properties \(\log_a⁡P-\log_a⁡Q=\log_a⁡({P \over Q})\) and \(\log_a⁡P^b=b \log_a⁡P\)

\(\lg⁡({10x \over x-9})=\lg⁡10^2\)

\(\lg⁡({10x \over x-9})=\lg⁡100\)

• Use the fact that \(\log_a⁡x=\log_a⁡y⟹x=y\)

\({10x \over x-9}=100\)

\(10x=100x-900\)

\(900=90x\)

\(10=x\)

• Ensure the validity of your answer.

---

LESSON 5

Solve the equation \(\log_2⁡x+4 \log_x⁡2=5\)

SOLUTION

\(\log_2⁡x+4 \log_x⁡2=5\)

• Use the property \(\log_a⁡x={1 \over \log_x⁡a}\)

\(\log_2⁡x+4({1 \over \log_2⁡x})=5\)

• Let \(y=\log_2⁡x\)

\(y+{4 \over y}=5\) \((×y)\)

\(y^2+4=5y\)

\(y^2-5y+4=0\)

\((y-1)(y-4)=0\)

\(y=1,4\)

Solve for \(x\)

\(y=1\)

\(⟹\log_2⁡x=1\)

Rewrite using indices

\(⟹x=2^1=2\)

\(y=4\)

\(⟹\log_2⁡x=4\)

\(⟹x=2^4=16\)

---

LESSON 6

Solve the equation \(\log_9⁡x=1+\log_3⁡3x, x>0\)

SOLUTION

• We need to have logarithms to the same base. We will change \(\log_9⁡x\) to base 3 but changing \(\log_3⁡3x\) to base 9 is also an option.

\(\log_9⁡x={\log_3⁡x \over \log_3⁡9}\)

\(={\log_3⁡x \over 2}\)

• Rewrite original equation

\(\log_9⁡x=1+\log_3⁡3x\)

\({1 \over 2 \log_3⁡x}=1+\log_3⁡3x\)

Convert 1 to \(\log_3⁡3\)

\({1 \over 2 \log_3⁡x}=\log_3⁡3+\log_3⁡3x\)

• Use the appropriate properties of logarithms to combine logarithms on the right hand side.

\(\log_3⁡x=2 \log_3⁡3+2 \log_3⁡3x\)

\(\log_3⁡x=\log_3⁡9+\log_3⁡ (9x^2)\)

\(\log_3⁡x=\log_3⁡(81x^2)\)

• Use the fact that \(\log_a⁡x=\log_a⁡y⟹x=y\)

\(x=81x^2\)

\(81x^2-x=0\)

\(x(81x-1)=0\)

\(x={1 \over 81}\) since \(x>0\)