Easy Logarithm Questions! Help?

1.
log₃10z
= log₃10 + log₃z

2.
log₂(√x y⁴ / z⁴)
= log₂√x + log₂y⁴ - log₂z⁴
= (1/2) log₂x + 4 log₂y - 4 log₂z

3.
log₄Z - log₄y
= log₄(Z/y)

4.
3 log₃x + 4 log₃y
= log₃x³ + log₃y⁴
= log₃(x³y⁴)

5.
4 eˣ = 91
logₑ(4 eˣ) = logₑ91
logₑ4 + logₑeˣ = logₑ91
x logₑeˣ = logₑ91 - logₑ4
x = logₑ91 - logₑ4
x = 3.1246

6.
(5) 10ˣ⁻⁶ = 7
10ˣ⁻⁶ = 7/5
log10ˣ⁻⁶ = log1.4
x - 6 = log1.4
x = 6 + log1.4
x = 6.1461

7.
8 (4⁶⁻²ˣ) + 13 = 41
8 (4⁶⁻²ˣ) = 28
4⁶⁻²ˣ = 3.5
log4⁶⁻²ˣ = log3.5
(6 - 2x) log4 = log3.5
6 - 2x = log3.5/log4
2x = 6 - (log3.5/log4)
x = (1/2) [6 - (log3.5/log4)]
x = 2.5482

8.
119 / (e⁶ˣ - 14) = 7
e⁶ˣ - 14 = 119/7
e⁶ˣ - 14 = 17
e⁶ˣ = 31
logₑe⁶ˣ = logₑ31
6x = logₑ31
x = (logₑ31)/6
x = 0.5723

9.
log(3z) = 2
3z = 10²
3z = 100
z = 33.3333

10.
5 * log(x - 2) = 11
log(x - 2) = 2.2
x - 2 = 10²˙²
x = 2 + 10²˙²
x = 160.4893

11.
7 + 3 * ln(x) = 5
3 ln(x) = -2
ln(x) = -2/3
x = e^(-2/3)
x = 0.5134

12.
log₂x + log₂(x + 2) = log₂(x + 6)
log₂[x(x + 2)] = log₂(x + 6)
x(x + 2) = x + 6
x² + 2x = x + 6
x² + x - 6 = 0
(x - 2)(x + 3) = 0
x = 2 or x = -3 (rejected, for log₂(-3) is undefined.)

Question1 depends on whether you mean log₃10z or log₃(10z).

log₃10z = z·log₃10 = log₃(10^z)
log₃(10z) = log₃(10)+log₃(z)
:::::
log₂(√xy⁴/z⁴) = log₂(√x) + log₂(y⁴) - log₂(z⁴)
= log₂(x^(½)) + log₂(y⁴) - log₂(z⁴)
= (½)log₂(x) + 4log₂(y) - 4log₂(z)

= 10.Log[3](10z) → you know that: Log[a](x) = Ln(x)/Ln(a) ← where a is the base

= 10.[Ln(10z) / Ln(3)] → you know that: Ln(ab) = Ln(a) + Ln(b)

= 10.[Ln(10) + Ln(z)] / Ln(3) → you know that: Ln(10) = Ln(2 * 5) = Ln(2) + Ln(5)

= 10.[Ln(2) + Ln(5) + Ln(z)] / Ln(3)


= 2.Log[2]{(√x * y⁴)/z⁴} → recall: Log[a](x) = Ln(x)/Ln(a) ← where a is the base

= 2.Ln{(√x * y⁴)/z⁴} / Ln(2) → you know that: Ln(a/b) = Ln(a) - Ln(b)

= 2.[Ln(√x * y⁴) - Ln(z⁴)] / Ln(2) → recall: Ln(ab) = Ln(a) + Ln(b)

= 2.[Ln(√x) + Ln(y⁴) - Ln(z⁴)] / Ln(2) → you know that: Ln(x^a) = a.Ln(x)

= 2.[Ln(√x) + 4.Ln(y) - 4.Ln(z)] / Ln(2) → you know that: √x = x^(1/2)

= 2.[Ln(x)^(1/2) + 4.Ln(y) - 4.Ln(z)] / Ln(2) → recall: Ln(x^a) = a.Ln(x)

= 2.[(1/2).Ln(x) + 4.Ln(y) - 4.Ln(z)] / Ln(2)

= [Ln(x) + 8.Ln(y) - 8.Ln(z)] / Ln(2)


= Log[4](z) - Log[4](y)

= [Ln(z) / Ln(4)] - [Ln(y) / Ln(4)]

= [Ln(z) - Ln(y)] / Ln(4)

= Ln(z/y) / Ln(4)

= Log[4)(z/y)


= 3.Log[3](x) + 4.Log[3](y)

= 3.[Ln(x) / Ln(3)] + 4.[Ln(y) / Ln(3)]

= [3.Ln(x) / Ln(3)] + [4.Ln(y) / Ln(3)]

= [3.Ln(x) + 4.Ln(y)] / Ln(3)

= [Ln(x³) + Ln(y⁴)] / Ln(3)

= Ln(x³.y⁴) / Ln(3)

= Log[3](x³.y⁴)


4.e^(x) = 91

e^(x) = 91/4

Ln[e^(x)] = Ln(91/4)

x.Ln(e) = Ln(91/4) → recall: Ln(e) = 1

x = Ln(91/4)

x = Ln(91) - Ln(4)


5/[10^(x - 6)] = 7

5/7 = 10^(x - 6)

Ln(5/7) = Ln[10^(x - 6)]

Ln(5/7) = (x - 6).Ln(10)

x - 6 = Ln(5/7) / Ln(10)

x = [Ln(5/7) / Ln(10)] + 6

x = [Ln(5/7) + 6.Ln(10)] / Ln(10)

x = [Ln(5) - Ln(7) + 6.Ln(2 * 5)] / Ln(10)

x = [Ln(5) - Ln(7) + 6.Ln(2) + 6.Ln(5)] / Ln(10)

x = [7.Ln(5) + 6.Ln(2) - Ln(7)] / Ln(10)


8^[4^(6 - 2x)] + 13 = 41

8^[4^(6 - 2x)] = 28

Ln{8^[4^(6 - 2x)]} = Ln(28)

4.(6 - 2x).Ln(8) = Ln(28)

6 - 2x = Ln(28) / [4.Ln(8)]

2x = 6 - { Ln(28) / [4.Ln(8)] }

x = 3 - { Ln(28) / [8.Ln(8)] }

x = [24.Ln(8) - Ln(28)] / [8.Ln(8)] → you know that : Ln(28) = Ln(4 * 7) = Ln(4) + Ln(7)

x = [24.Ln(8) - Ln(4) - Ln(7)] / [8.Ln(8)] → you know that : Ln(8) = Ln(2³) = 3.Ln(2)

x = [72.Ln(2) - Ln(4) - Ln(7)] / [24.Ln(2)] → you know that : Ln(4) = Ln(2²) = 2.Ln(2)

x = [72.Ln(2) - 2.Ln(2) - Ln(7)] / [24.Ln(2)]

x = [70.Ln(2) - Ln(7)] / [24.Ln(2)]


119/[e^(6x) - 14] = 7

119/7 = e^(6x) - 14

17 = e^(6x) - 14

e^(6x) = 17 + 14

e^(6x) = 31

Ln[e^(6x)] = Ln(31)

6x.Ln(e) = Ln(31) → recall: Ln(e) = 1

6x = Ln(31)

x = (1/6).Ln(31)


Log(3z) = 2 → you know that: Log(x) = Ln(x) / Ln(10)

Ln(3z) / Ln(10) = 2

Ln(3z) = 2.Ln(10)

Ln(3z) = Ln(10²)

3z = 100

z = 100/3


5.Log(x - 2) = 11

Log(x - 2) = 11/5

Ln(x - 2) / Ln(10) = 11/5

Ln(x - 2) = (11/5).Ln(10)

Ln(x - 2) = Ln(10)^(11/5)

x - 2 = 10^(11/5)

x = 2 + 10^(11/5)


7 + 3.Ln(x) = 5

3.Ln(x) = - 2

Ln(x) = - 2/3

e^[Ln(x)] = e^(- 2/3)

x = e^(- 2/3)


Log[2](x) + Log[2](x + 2) = Log[2](x + 6)


First condition : x > 0

Second condition: (x + 2) > 0 → x + 2 > 0 → x > - 2

Third condition: (x + 6) > 0 → x + 6 > 0 → x > - 6

You join the 3 conditions and you can obtain only one: x > 0


[Ln(x) / Ln(2)] + [Ln(x + 2) / Ln(2)] = [Ln(x + 6) / Ln(2)]

Ln(x) + Ln(x + 2) = Ln(x + 6)

Ln[x.(x + 2)] = Ln(x + 6)

x.(x + 2) = x + 6

x² + 2x = x + 6

x² + x = 6

x² + x + (1/2)² = 6 + (1/2)²

x² + x + (1/2)² = 25/4

[x + (1/2)]² = (± 5/2)²

x + (1/2) = ± 5/2

x = - (1/2) ± (5/2)

x = (- 1 ± 5)/2


x = (- 1 - 5)/2 → x = - 6/2 → x = - 3 → no possible because the condition

x = (- 1 + 5)/2 → x = 4/2 → x = 2

Use Log rules
1/
log10+logz
2/
(1/2)logx+4logy-4logz
3/
log(z/y)
4/
log(x^3*y^4)
5/
x=ln(91/4)=...calculator
6/
10^(x-6)=7/5
x-6 = log(7/5)
x=6 + log(7/5) = ...calculator