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2 years ago

8

Help please logarithms

1 answer:

DochEvi [55]2 years ago

6 0

Answer:

Simplified: 40

Step-by-step explanation:

To simplify this, we need to first separate the terms in this expression. There is a property of exponents that says: $x^{a+b} =x^ax^b$ We can do this same thing to simplify this expression. Next, we can simplify each of these terms. $2^2$ simply becomes 4. In the other term, the $2^{log_2}$ simply cancel each other and leave the 10. This means that we're left with 4*10, which is 40

$2^{2+log_210} \\\\2^2*2^{log_210} \\\\4*10\\\\40$

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3 years ago

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2 years ago

Use algebra to find the point at which the line h(x)=−(3/5)x+(7/5) intersects the line k(x)=(−8/3)x+(404/45).

Inga [223]

Answer: the point (3.667, 0.8)

Step-by-step explanation:

We want to find the point at which the lines:

h(x) = -(3/5)*x + (7/5)

and

k(x) = (-8/3)*x + (404/45)

If the lines intersect, then we must have:

h(x) = k(x)

if we write the functions we get:

-(3/5)*x + (7/5) = (-8/3)*x + (404/45)

Now we need to solve this for x.

-(3/5)*x + (8/3)*x = (404/45) - (7/5)

(8/3 - 3/5)*x = 404/45 - 63/45 = 341/45

(40/15 - 9/15)*x = 341/45

(31/15)*x = 341/45

x = (341/45)*(15/31) = 3.667

Now we can input this value of x in the functions to get the output.

h(3.667) = -(3/5)*3.667 + 7/5 = -0.8

Then the point where our points intersect is the point (3.667, 0.8)

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2 years ago