Question

Please help! Its for my big test tomorrow!

Answer 1

QUESTION 1

`{3}^(x + 1) = {9}^(x + 3)`

This is the same as:

`{3}^(x + 1) = {3}^(2(x + 3))`

Equate the exponents.

x+1=2(x+3)

Expand:

x+1=2x+6

Group similar terms;

2x-x=1-6

x=-5

QUESTION 2

`log(9x - 2) = log(4x + 3)`

Equate the arguments.

9x-2=4x+3

Group similar terms;

9x-4x=3+2

5x=5

Divide through by 5

x=1

QUESTION 3

`log_(6)(5x + 4) = 2`

Take antilogarithm to obtain,

`5x + 4 = {6}^(2)`

This implies that,

5x+4=36

5x=36-4

5x=32

x=32/5

or

`x = 6 (2)/(5)`

QUESTION 4

`log_(2)(x) + log_(2)(x - 3) = 2`

Use the product rule of logarithms:

`log_(2)x(x - 3) = 2`

Take antilogarithm,

`{x}^(2) - 3x = {2}^(2)`

`{x}^(2) - 3x - 4 = 0`

Factor:

`(x + 1)(x - 4) = 0`

This implies that,

`x = - 1 \: or \: x = 4`

But the domain is x>0, therefore the solution is

x=4

QUESTION 5

`x=\log_(4)(11.2)`

`x=\log_(4)((56)/(5))`

`x=\log_(4)(56)-\log_(4)(5)`

x=1.7 to the nearest tenth.

QUESTION 6

`2e^(8x)=9.2`

Divide both sides by 2.

`e^(8x)=4.6`

Take natural log of both sides

`{8x}=\ln(4.6)`

`{x}=\ln(4.6)/ 8`

x=0.2 to the nearest tenth.

Answer 2

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m × b^n = b^(m + n) ⇒ in multiplication if they have same base

we add the power

2- b^m ÷ b^n = b^(m – n) ⇒ in division if they have same base we

subtract the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

them

4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same

power then we multiply them ad put over the answer the power

5- b^(-m) = 1/(b^m) (for all nonzero real numbers b) ⇒ If we have

negative power we reciprocal the base to get positive power

6- If a^m = a^n , then m = n ⇒ equal bases get equal powers

7- If a^m = b^m , then a = b or m = 0

# Logarithmic rules:

1-
`log_(a)b=n-----a^(n)=b`

2-
`loga_(1)=0---log_(a)a=1---ln(e)=1`

3-
`log_(a)q+log_(a)p=log_(a)qp`

4-
`log_(a)q-log_(a)p=log_(a)(q)/(p)`

5-
`log_(a)q^(n)=nlog_(a)q`

* Now lets solve the problems

#
`3^(x+1)=9^(x+3)`

- Change the base 9 to 3²

∴
`9^(x+3)=3^(2(x+3))=3^(2x+6)`

∴
`3^(x+1)=3^(2x+6)`

- Same bases have equal powers

∴ x + 1 = 2x + 6 ⇒ subtract x and 6 from both sides

∴ 1 - 6 = 2x - x

∴ -5 = x

* The solution x = -5

# ㏒(9x - 2) = ㏒(4x + 3)

- If ㏒(a) = ㏒(b), then a = b

∴ 9x - 2 = 4x + 3 ⇒ subtract 4x from both sides and add 2 to both sides

∴ 5x = 5 ⇒ divide both sides by 5

∴ x = 1

* The solution is x = 1

#
`log_(6)(5x+4)=2`

- Use the 1st rule in the logarithmic equation

∴ 6² = 5x + 4

∴ 36 = 5x + 4 ⇒ subtract 4 from both sides

∴ 32 = 5x ⇒ divide both sides by 5

∴ 6.4 = x

* The solution is x = 6.4

#
`log_(2)x+log_(2)(x-3)=2`

- Use the rule 3 in the logarithmic equation

∴
`log_(2)x(x-3)=2`

- Use the 1st rule in the logarithmic equation

∴ 2² = x(x - 3) ⇒ simplify

∴ 4 = x² - 3x ⇒ subtract 4 from both sides

∴ x² - 3x - 4 = 0 ⇒ factorize it into two brackets

∴ (x - 4)(x + 1) = 0 ⇒ equate each bract by 0

∴ x - 4 = 0 ⇒ add 4 to both sides

∴ x = 4

OR

∵ x + 1 = 0 ⇒ subtract 1 from both sides

∴ x = -1

- We will reject this answer because when we substitute the value

of x in the given equation we will find
`log_(2)(-1)` and this

value is undefined, there is no logarithm for negative number

* The solution is x = 4

#
`log_(4)11.2=x`

- You can use the calculator directly to find x

∴ x = 1.7427

* The solution is 1.7427

#
`2e^(8x)=9.2` ⇒ divide the both sides by 2

∴
`e^(8x)=4.6`

- Insert ln for both sides

∴
`lne^(8x)=ln(4.6)`

- Use the rule
`ln(e^(n))=nln(e)` ⇒ ln(e) = 1

∴ 8x = ln(4.6) ⇒ divide both sides by 8

∴ x = ln(4.6)/8 = 0.190757

* The solution is 0.190757