$Domain:\ x > 0\\\\3\ln(x)+2\ln(4)=\ln(128)\qquad\text{use}\ \lag_ab^n=n\log_ab\\\\\ln(x^3)+\ln(4^2)=\ln(128)\qquad\text{use}\ \log_ab+\ln_ac=\ln_a(bc)\\\\\ln(16x^3)=\ln(128)\iff16x^3=128\qquad\text{divide both sides by 16}\\\\x^3=8\to x=\sqrt[3]8\\\\\boxed{x=2}\in D$

3 0

2 years ago

You purchase 5 pounds of apples and 2 pounds of oranges for $9.Your friend purchases 5 pounds of apples and 6 pounds of oranges

babymother [125]

Answer:

x = 1 and y = 2

Step-by-step explanation:

Let apples are represented by x

and let oranges are represented by y

You purchase 5 pounds of apples and 2 pounds of oranges for $9. This line in equation format can be written as:

5x + 2y = 9

Your friend purchases 5 pounds of apples and 6 pounds of oranges for $17.

This line in equation format can be written as:

5x + 6y = 17

Now we have two equations:

5x + 2y = 9 -> eq (i)

5x + 6y = 17 -> eq(ii)

We can solve these equations to find the value of x and y.

Subtracting eq(i) from eq(ii)

5x + 6y = 17

5x + 2y = 9

- - -

_________

0+4y= 8

=> 4y = 8

y= 8/4

y = 2

Now, putting value of y in eq (i)

5x + 2y = 9

5x +2(2) = 9

5x +4 = 9

5x = 9-4

5x = 5

x = 1

so, x = 1 and y = 2

8 0

2 years ago