All categories

3 years ago

5^{x+1}-5^{x+2}=-2500. Find x

Mathematics

1 answer:

vladimir1956 [14]3 years ago

4 0

Answer:

x is 3

Step-by-step explanation:

${5}^{(x + 1)} - {5}^{(x + 2)} = - 2500 \\$

divide all sides by -1:

${5}^{(x + 2)} - 5 {}^{(x + 1)} = 2500$

from law of indices:

$( {5}^{x} . {5}^{2} ) - ( {5}^{x} . {5}^{1} ) = 2500$

factorise out 5^x:

${5}^{x} ( {5}^{2} - {5}^{1} ) = 2500 \\ {5}^{x} (20) = 2500 \\ {5}^{x} = 125 \\ {5}^{x} = {5}^{3} \\ x = 3$

You might be interested in

My last one got deleted but whats 10x5 i dont actually but u get it

andre [41]

Answer:

50 and 160

Step-by-step explanation:

good

6 0

3 years ago

Read 2 more answers

300 30 3 0.3 follow pattern

natita [175]

.03 .003 .0003....etc

5 0

3 years ago

Read 2 more answers

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.