QUESTION 1
[tex] {3}^{x + 1} = {9}^{x + 3} [/tex]
This is the same as:
[tex] {3}^{x + 1} = {3}^{2(x + 3)} [/tex]
Equate the exponents.
x+1=2(x+3)
Expand:
x+1=2x+6
Group similar terms;
2x-x=1-6
x=-5
QUESTION 2
[tex] log(9x - 2) = log(4x + 3) [/tex]
Equate the arguments.
9x-2=4x+3
Group similar terms;
9x-4x=3+2
5x=5
Divide through by 5
x=1
QUESTION 3
[tex] log_{6}(5x + 4) = 2[/tex]
Take antilogarithm to obtain,
[tex]5x + 4 = {6}^{2} [/tex]
This implies that,
5x+4=36
5x=36-4
5x=32
x=32/5
or
[tex]x = 6 \frac{2}{5} [/tex]
QUESTION 4
[tex] log_{2}(x) + log_{2}(x - 3) = 2[/tex]
Use the product rule of logarithms:
[tex]log_{2}x(x - 3) = 2[/tex]
Take antilogarithm,
[tex] {x}^{2} - 3x = {2}^{2} [/tex]
[tex] {x}^{2} - 3x - 4 = 0[/tex]
Factor:
[tex](x + 1)(x - 4) = 0[/tex]
This implies that,
[tex]x = - 1 \: or \: x = 4[/tex]
But the domain is x>0, therefore the solution is
x=4
QUESTION 5
[tex]x=\log_{4}(11.2)[/tex]
[tex]x=\log_{4}(\frac{56}{5})[/tex]
[tex]x=\log_{4}(56)-\log_{4}(5)[/tex]
x=1.7 to the nearest tenth.
QUESTION 6
[tex]2e^{8x}=9.2[/tex]
Divide both sides by 2.
[tex]e^{8x}=4.6[/tex]
Take natural log of both sides
[tex]{8x}=\ln(4.6)[/tex]
[tex]{x}=\ln(4.6)\div 8[/tex]
x=0.2 to the nearest tenth.
Answer:
# 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
Step-by-step 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- [tex]log_{a}b=n-----a^{n}=b[/tex]
2- [tex]loga_{1}=0---log_{a}a=1---ln(e)=1[/tex]
3- [tex]log_{a}q+log_{a}p=log_{a}qp[/tex]
4- [tex]log_{a}q-log_{a}p=log_{a}\frac{q}{p}[/tex]
5- [tex]log_{a}q^{n}=nlog_{a}q[/tex]
* Now lets solve the problems
# [tex]3^{x+1}=9^{x+3}[/tex]
- Change the base 9 to 3²
∴ [tex]9^{x+3}=3^{2(x+3)}=3^{2x+6}[/tex]
∴ [tex]3^{x+1}=3^{2x+6}[/tex]
- 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
# [tex]log_{6}(5x+4)=2[/tex]
- 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
# [tex]log_{2}x+log_{2}(x-3)=2[/tex]
- Use the rule 3 in the logarithmic equation
∴ [tex]log_{2}x(x-3)=2[/tex]
- 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 [tex]log_{2}(-1)[/tex] and this
value is undefined, there is no logarithm for negative number
* The solution is x = 4
# [tex]log_{4}11.2=x[/tex]
- You can use the calculator directly to find x
∴ x = 1.7427
* The solution is 1.7427
# [tex]2e^{8x}=9.2[/tex] ⇒ divide the both sides by 2
∴ [tex]e^{8x}=4.6[/tex]
- Insert ln for both sides
∴ [tex]lne^{8x}=ln(4.6)[/tex]
- Use the rule [tex]ln(e^{n})=nln(e)[/tex] ⇒ ln(e) = 1
∴ 8x = ln(4.6) ⇒ divide both sides by 8
∴ x = ln(4.6)/8 = 0.190757
* The solution is 0.190757
