Notes:

1. a log b = log b^a

2. log b + log c = log b x c

3. log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log x³ + log y - log z²

3 log x + log y - 2 log z = log x³ x y - log z²

3 log x + log y - 2 log z = log x³y - log z²

Apply log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log ($\frac{{\mathrm{x}}^3\mathrm{y}}{{\mathrm{z}}^2}$)

log x = 2 - 3log 2

Group the logs

log x + 3log 2 = 2

From law of logarithms

nlogm = logmⁿ

3log 2 = log 2³

log x + log 2³ = 2

From law of logarithms

LogA + LogB = Log A x B

log (x x 2³) = 2

2³ = 2 x 2 x 2 = 8

log (x x 8) = 2

log 8x = 2

From law of logarithms

log _a^b = c → a^c = b

Where a is the base and b is the logarithm

The base of the logarithm log 8x is 10

10² = 8x

10² = 10 x 10 = 100

100 = 8x

Divide both sides by 8

x = $\frac{100}{8}$ = $\frac{25}{2}$

Note: 4 divides 100, 25 times and 8, 2 times.

logb^c = clogb

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

6log(x + 4) = log64

6log(x + 4) = log2⁶
6log(x + 4) = 6log2

Compare the left side and the right side.

x + 4 = 2
x = 2 - 4
x = -2

24 = 8 x 3

8 = 2 x 2 x 2 = 2³

24 = 2³ x 3

log₁₀ ²⁴ = log₁₀ 2³ x 3

From law of logarithm,

1. log a^b = blog a

2. log a x b = log a + log b

log₁₀ 2³ x 3 = 3log₁₀ 2 x 3

3log₁₀ 2 x 3 = 3log₁₀ 2 + log₁₀ 3

But log₁₀ 2 = m and log₁₀ 3 = n

3log₁₀ 2 + log₁₀ 3 = 3 x m + n

3log₁₀ 2 + log₁₀ 3 = 3m + n

Notes

1. log_bb = 1

log₁₀10 = 1

2. blog c = log c^b

2log₁₀3 = log₁₀3²

3. log b + log c = log b x c

1 + 2 log₁₀3 = log₁₀10 + log₁₀3²

1 + 2 log₁₀3 = log₁₀10 + log₁₀9

1 + 2 log₁₀3 = log₁₀10 x 9

1 + 2 log₁₀3 = log₁₀90

9 = 3²

log₃9 = log₃3²

log^ba = bloga

log₃3² = 2log₃3

log_aa = 1

log₃3 = 1

2log₃3 = 2 x 1 = 2

∴ log₃9 = 2

8 = 2³

log₂8 = log₂2³

log₂2³ = 3log₂2

log₂2 = 1

log₂2³ = 3 x 1 = 3

∴ log₂8 = 3

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27

27 = 3³

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀3³

loga^b = bloga

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+3 x $\frac{2}{3}$log₁₀3

Note:

3 x $\frac{2}{3}$ = 2

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+2log₁₀3

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-log₁₀3

loga - logb = log$\frac{\mathrm{a}}{\mathrm{b}}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀$\frac{6}{3}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀2

log_x8 = log_x8

8 = 2³

log_x8 = log_x2³

loga^b = bloga

log_x8 = 3log_x2

But log_x2 = 0.3

log_x8 = 3 x 0.3

log_x8 = 0.9

Notes:

1. log a^b = b log a

2. ${\mathrm{log\; Z}}^{\frac{1}{3}}$ = $\frac{1}{3}$ x log Z

3. log y³ = 3 log y

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{log\; x\; -\; }\frac{1}{3}\mathrm{log\; Z}}{\mathrm{3log\; y}}$

log x = 0.3030, log y = 0.4777 and log Z = 0.8451

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{0.3030\; -\; }\frac{1}{3}\mathrm{x\; 0.8451}}{\mathrm{3\; x\; 0.4777}}$

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = 0.01486 ≈ 0.0149

Applying log_a^bc = c x log_a^b = clog_a^b

log₁₀^2y = y x log₁₀²

But log₁₀^2y = 1.8060

y x log₁₀² = 1.8060

But log₁₀² = 0.3010

Hence y x 0.3010 = 1.8060

Divide both sides by 0.3010

y = $\frac{1.8060}{0.3010}$ = 6

$\frac{log{3}^{(2x\; -\; 1)}}{log{3}^{243}}$ = $\frac{2}{5}$

243 = 3⁵

log_a^ab = b x log_a^a

But log_a^a = 1

log_a^ab = b x 1

log_a^ab = b

log₃²⁴³ = log₃³⁵

log₃³⁵ = 5

$\frac{log{3}^{(2x\; -\; 1)}}{5}$ = $\frac{2}{5}$

Since the denominator at the left equals to the denominator at the right, the numerators are also the same.

log₃^{(2x - 1)} = 2

If log_b^a = c, b^c = a

3² = 2x - 1

9 = 2x - 1

9 + 1 = 2x

10 = 2x

Divide both sides by 2

x = $\frac{10}{2}$

x = 5

Notes:

1. a log b = log b^a

2. log b + log c = log b x c

3. log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log x³ + log y - log z²

3 log x + log y - 2 log z = log x³ x y - log z²

3 log x + log y - 2 log z = log x³y - log z²

Apply log b - log c = log $\frac{\mathrm{b}}{\mathrm{c}}$

3 log x + log y - 2 log z = log ($\frac{{\mathrm{x}}^3\mathrm{y}}{{\mathrm{z}}^2}$)

log x = 2 - 3log 2

Group the logs

log x + 3log 2 = 2

From law of logarithms

nlogm = logmⁿ

3log 2 = log 2³

log x + log 2³ = 2

From law of logarithms

LogA + LogB = Log A x B

log (x x 2³) = 2

2³ = 2 x 2 x 2 = 8

log (x x 8) = 2

log 8x = 2

From law of logarithms

log _a^b = c → a^c = b

Where a is the base and b is the logarithm

The base of the logarithm log 8x is 10

10² = 8x

10² = 10 x 10 = 100

100 = 8x

Divide both sides by 8

x = $\frac{100}{8}$ = $\frac{25}{2}$

Note: 4 divides 100, 25 times and 8, 2 times.

logb^c = clogb

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

6log(x + 4) = log64

6log(x + 4) = log2⁶
6log(x + 4) = 6log2

Compare the left side and the right side.

x + 4 = 2
x = 2 - 4
x = -2

24 = 8 x 3

8 = 2 x 2 x 2 = 2³

24 = 2³ x 3

log₁₀ ²⁴ = log₁₀ 2³ x 3

From law of logarithm,

1. log a^b = blog a

2. log a x b = log a + log b

log₁₀ 2³ x 3 = 3log₁₀ 2 x 3

3log₁₀ 2 x 3 = 3log₁₀ 2 + log₁₀ 3

But log₁₀ 2 = m and log₁₀ 3 = n

3log₁₀ 2 + log₁₀ 3 = 3 x m + n

3log₁₀ 2 + log₁₀ 3 = 3m + n

Notes

1. log_bb = 1

log₁₀10 = 1

2. blog c = log c^b

2log₁₀3 = log₁₀3²

3. log b + log c = log b x c

1 + 2 log₁₀3 = log₁₀10 + log₁₀3²

1 + 2 log₁₀3 = log₁₀10 + log₁₀9

1 + 2 log₁₀3 = log₁₀10 x 9

1 + 2 log₁₀3 = log₁₀90

9 = 3²

log₃9 = log₃3²

log^ba = bloga

log₃3² = 2log₃3

log_aa = 1

log₃3 = 1

2log₃3 = 2 x 1 = 2

∴ log₃9 = 2

8 = 2³

log₂8 = log₂2³

log₂2³ = 3log₂2

log₂2 = 1

log₂2³ = 3 x 1 = 3

∴ log₂8 = 3

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27

27 = 3³

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀3³

loga^b = bloga

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+3 x $\frac{2}{3}$log₁₀3

Note:

3 x $\frac{2}{3}$ = 2

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-3log₁₀3+2log₁₀3

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀6-log₁₀3

loga - logb = log$\frac{\mathrm{a}}{\mathrm{b}}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀$\frac{6}{3}$

log₁₀6-3log₁₀3+$\frac{2}{3}$log₁₀27 = log₁₀2

log_x8 = log_x8

8 = 2³

log_x8 = log_x2³

loga^b = bloga

log_x8 = 3log_x2

But log_x2 = 0.3

log_x8 = 3 x 0.3

log_x8 = 0.9

Notes:

1. log a^b = b log a

2. ${\mathrm{log\; Z}}^{\frac{1}{3}}$ = $\frac{1}{3}$ x log Z

3. log y³ = 3 log y

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{log\; x\; -\; }\frac{1}{3}\mathrm{log\; Z}}{\mathrm{3log\; y}}$

log x = 0.3030, log y = 0.4777 and log Z = 0.8451

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = $\frac{\mathrm{0.3030\; -\; }\frac{1}{3}\mathrm{x\; 0.8451}}{\mathrm{3\; x\; 0.4777}}$

$\frac{\mathrm{log\; x\; -\; }{\mathrm{log\; Z}}^{\frac{1}{3}}}{{\mathrm{log\; y}}^3}$ = 0.01486 ≈ 0.0149

Applying log_a^bc = c x log_a^b = clog_a^b

log₁₀^2y = y x log₁₀²

But log₁₀^2y = 1.8060

y x log₁₀² = 1.8060

But log₁₀² = 0.3010

Hence y x 0.3010 = 1.8060

Divide both sides by 0.3010

y = $\frac{1.8060}{0.3010}$ = 6