Question

Please help! Its for my big test tomorrow!

Answer 1

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- $log_{a}b=n-----a^{n}=b$

2- $loga_{1}=0---log_{a}a=1---ln(e)=1$

3- $log_{a}q+log_{a}p=log_{a}qp$

4- $log_{a}q-log_{a}p=log_{a}\frac{q}{p}$

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

Answer 2

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

$x = 6 \frac{2}{5}$

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

$x=\log_{4}(11.2)$

$x=\log_{4}(\frac{56}{5})$

$x=\log_{4}(56)-\log_{4}(5)$

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

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

${x}=\ln(4.6)\div 8$

x=0.2 to the nearest tenth.