Intro to logarithms

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Intro to logarithms

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We have the following definition of logarithms:

Definition

For a > 0, a 6= 1 and b > 0 we have:

log_ab = c ⇔ a^c = b

What does it mean? First of all the assumptions (restrictions) are important. The number a, called the base of the logarithm, has to be greater than 0 and cannot be equal to 1. The number b (which we take the logarithm of) has to be greater than 0.

So the expressions like log₁3, log₋₂5 or log₄(−1) are not defined in real numbers (similarly to expressions like √

−6).

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Definition

log_ab = c ⇔ a^c = b What does it mean?

First of all the assumptions (restrictions) are important. The number a, called the base of the logarithm, has to be greater than 0 and cannot be equal to 1. The number b (which we take the logarithm of) has to be greater than 0.

−6).

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Definition

What does it mean? First of all the assumptions (restrictions) are important. The number a, called the base of the logarithm, has to be greater than 0 and cannot be equal to 1.

The number b (which we take the logarithm of) has to be greater than 0.

−6).

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Definition

−6).

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Definition

−6).

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Definition

Secondly log_ab = c means a raised to the power of c is equal to b. So if we want to calculate log_ab, we need to find a number to which we need to raise a to to get b.

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Definition

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Definition

Secondly log_ab = c means a raised to the power of c is equal to b.

So if we want to calculate log_ab, we need to find a number to which we need to raise a to to get b.

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Definition

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We will practice the above definition in this presentation.

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Example 1

Calculate log1 381.

We need to find the power to which to raise ¹₃ to get 81. This can be written as an exponential equation:

1 3

x

= 81 So we have:

3^−x = 3⁴ which gives x = −4.

So log1

381 = −4.

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Example 1

Calculate log1 381.

We need to find the power to which to raise ¹₃ to get 81.

This can be written as an exponential equation:

1 3

x

= 81 So we have:

So log1

381 = −4.

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Example 1

Calculate log1 381.

1 3

x

= 81

So we have:

So log1

381 = −4.

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Example 1

Calculate log1 381.

1 3

x

= 81 So we have:

3^−x = 3⁴

which gives x = −4. So log1

381 = −4.

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Example 1

Calculate log1 381.

1 3

x

= 81 So we have:

So log1

381 = −4.

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Example 1

Calculate log1 381.

1 3

x

= 81 So we have:

So log1

381 = −4.

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Example 2

Calculate log₆ ₂₁₆¹ .

We need to find the power to which to raise 6 to get ₂₁₆¹ . We can write this as an exponential eqaution:

6^x = 1 216 This gives:

6^x = 6⁻³ so x = −3.

So log₆₂₁₆¹ = −3.

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Example 2

We need to find the power to which to raise 6 to get ₂₁₆¹ .

We can write this as an exponential eqaution:

6^x = 6⁻³ so x = −3.

So log₆₂₁₆¹ = −3.

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Example 2

6^x = 1 216

This gives:

6^x = 6⁻³ so x = −3.

So log₆₂₁₆¹ = −3.

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Example 2

6^x = 6⁻³

so x = −3. So log₆₂₁₆¹ = −3.

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Example 2

6^x = 6⁻³ so x = −3.

So log₆₂₁₆¹ = −3.

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Example 2

6^x = 6⁻³ so x = −3.

So log₆₂₁₆¹ = −3.

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Example 3

Calculate log¹

416.

We need to find the power to which to raise ¹₄, to get 16. We can rewrite this as an exponential equation:

1 4

x

= 16 We can change all terms into powers of 4 (or 2)

4^−x = 4² so x = −2.

So log¹

416 = −2.

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Example 3

Calculate log¹

416.

We need to find the power to which to raise ¹₄, to get 16.

We can rewrite this as an exponential equation:

1 4

x

4^−x = 4² so x = −2.

So log¹

416 = −2.

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Example 3

Calculate log¹

416.

1 4

x

= 16

We can change all terms into powers of 4 (or 2) 4^−x = 4² so x = −2.

So log¹

416 = −2.

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Example 3

Calculate log¹

416.

1 4

x

4^−x = 4²

so x = −2. So log¹

416 = −2.

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Example 3

Calculate log¹

416.

1 4

x

4^−x = 4² so x = −2.

So log¹

416 = −2.

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Example 3

Calculate log¹

416.

1 4

x

4^−x = 4² so x = −2.

So log¹16 = −2.

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Example 4

Calculate log₂^√₂16.

We need to find the power to which to raise 2√

2 to get 16. We write the corresponding exponential equation:

(2√

2)^x = 16 Change into powers of 2:

2³²^x = 2⁴ so x = ⁸₃.

So we have log₂^√₂16 = ⁸₃.

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Example 4

2 to get 16.

We write the corresponding exponential equation:

(2√

2³²^x = 2⁴ so x = ⁸₃.

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Example 4

(2√

2)^x = 16

Change into powers of 2:

2³²^x = 2⁴ so x = ⁸₃.

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Example 4

(2√

2³²^x = 2⁴

so x = ⁸₃.

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Example 4

(2√

2³²^x = 2⁴ so x = ⁸₃.

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Example 4

(2√

2³²^x = 2⁴ so x = ⁸₃.

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Example 5

Calculate log₅125√ 5.

We need to find the power to which to raise 5 to get 125√ 5. The corresponding exponential equation is:

5^x = 125√ 5 Which gives:

5^x = 5^3.5 so x = 3.5.

So we have log₅125√

5 = 3.5.

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Example 5

We need to find the power to which to raise 5 to get 125√ 5.

The corresponding exponential equation is:

5^x = 5^3.5 so x = 3.5.

5 = 3.5.

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Example 5

5^x = 125√ 5

Which gives:

5^x = 5^3.5 so x = 3.5.

5 = 3.5.

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Example 5

5^x = 5^3.5

so x = 3.5.

5 = 3.5.

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Example 5

5^x = 5^3.5 so x = 3.5.

5 = 3.5.

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Example 5

5^x = 5^3.5 so x = 3.5.

5 = 3.5.

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Example 6

Calculate log₃^√₃81√³ 3.

3 to get 81√³

3. We need to solve the following exponential equation:

(3√

3)^x = 81√³ 3 Change into powers of 3:

3³²^x = 3⁴¹³ this gives x = ²⁶₉.

So we have log₃^√₃81√³

3 = ²⁶₉.

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Example 6

3 to get 81√³ 3.

We need to solve the following exponential equation:

(3√

3³²^x = 3⁴¹³ this gives x = ²⁶₉.

3 = ²⁶₉.

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Example 6

3 to get 81√³

(3√

3)^x = 81√³ 3

Change into powers of 3:

3³²^x = 3⁴¹³ this gives x = ²⁶₉.

3 = ²⁶₉.

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Example 6

3 to get 81√³

(3√

3³²^x = 3⁴¹³

this gives x = ²⁶₉. So we have log₃^√₃81√³

3 = ²⁶₉.

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Example 6

3 to get 81√³

(3√

3³²^x = 3⁴¹³ this gives x = ²⁶₉.

3 = ²⁶₉.

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Example 6

3 to get 81√³

(3√

3³²^x = 3⁴¹³ this gives x = ²⁶₉.

3 = ²⁶₉.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

We need to find the power to which to raise ¹₄ to get ²⁵

√

√64

8 . We write the corresponding exponential equation:

1 4

x

= 2√⁵

√64 8 Change into powers of 2

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷ which gives x = −₂₀⁷ .

So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64 8 .

We write the corresponding exponential equation:

1 4

x

= 2√⁵

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷ which gives x = −₂₀⁷ .

So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64

1 4

x

= 2√⁵

√64 8

Change into powers of 2

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷ which gives x = −₂₀⁷ .

So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64

1 4

x

= 2√⁵

2^−2x = 2 · 2⁶⁵ 2³²

so:

2^−2x = 2¹⁰⁷ which gives x = −₂₀⁷ .

So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64

1 4

x

= 2√⁵

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷

which gives x = −₂₀⁷ . So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64

1 4

x

= 2√⁵

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷ which gives x = −₂₀⁷ .

So in the end log¹

4

2√⁵

√64

8 = −₂₀⁷.

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Example 7

Calculate log¹

4

2√⁵

√64 8 .

√

√64

1 4

x

= 2√⁵

2^−2x = 2 · 2⁶⁵ 2³² so:

2^−2x = 2¹⁰⁷

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Example 8

Solve the following equation:

log₂x = −3

Note that we must have x > 0. The corresponding exponential equation is:

2⁻³ = x So x = 1

8.

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Example 8

log₂x = −3

Note that we must have x > 0.

2⁻³ = x So x = 1

8.

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Example 8

log₂x = −3

2⁻³ = x So x = 1

8.

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Example 8

log₂x = −3

2⁻³ = x So x = 1

8.

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Example 8

log₂x = −3

2⁻³ = x

So x = 1 8.

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Example 8

log₂x = −3

2⁻³ = x So x = 1

8.

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Example 9

log^√₃x = −4

Note that again we must have x > 0. The corresponding exponential equation is:

(√

3)⁻⁴= x So x = 1

9.

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Example 9

log^√₃x = −4

Note that again we must have x > 0.

(√

3)⁻⁴= x So x = 1

9.

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Example 9

log^√₃x = −4

(√

3)⁻⁴= x So x = 1

9.

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Example 9

log^√₃x = −4

(√

3)⁻⁴= x So x = 1

9.

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Example 9

log^√₃x = −4

(

√

3)⁻⁴ = x

So x = 1 9.

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Example 9

log^√₃x = −4

(

√

3)⁻⁴ = x So x = 1

.

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Example 10

log_x9 = 2

Note that we must have x > 0 and x 6= 1. The corresponding exponential equation is:

x² = 9 So x = 3 (x 6= −3, as it has to be positive).

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Example 10

log_x9 = 2

Note that we must have x > 0 and x 6= 1.

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Example 10

log_x9 = 2

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Example 10

log_x9 = 2

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Example 10

log_x9 = 2

x² = 9

So x = 3 (x 6= −3, as it has to be positive).

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Example 10

log_x9 = 2

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Example 11

log_x64 = 3

x³ = 64 So x = 4.

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Example 11

log_x64 = 3

Note that we must have x > 0 and x 6= 1.

x³ = 64 So x = 4.

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Example 11

log_x64 = 3

x³ = 64 So x = 4.

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Example 11

log_x64 = 3

x³ = 64 So x = 4.

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Example 11

log_x64 = 3

x³ = 64

So x = 4.

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Example 11

log_x64 = 3

x³ = 64 So x = 4.

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In case of any questions you can email me at t.j.lechowski@gmail.com.