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4 years ago

If a^16 is the same as 2^64, what is the absolute value of a?

Mathematics

1 answer:

Alisiya [41]4 years ago

3 0

Answer:

$a=16$

Step-by-step explanation:

we know the power rule:

$(a^{m})^{n} =a^{m*n}$

In this problem we have

$a^{16}=2^{64}$

rasise both sides to 1/16

$(a^{16})^{\frac{1}{16}}=(2^{64})^{\frac{1}{16}}$

Applying the power rule

$(a)^{\frac{16}{16}}=(2)^{\frac{64}{16}$

simplify

$a=(2)^{4}$

$a=16$

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Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago

How much would you pay today to receive $50 in one year and $60 in the second year if you can earn 15 percent interest on altern

Andrej [43]

The present value of an income of $60 in 2 years and $50 in 1 year at an interest rate of 15% can be computed as ...

... ($60/1.15 +$50)/1.15 = $88.85

_____

The present value (PV) of an amount (A) received 1 year from now earning annual interest rate r is ...

... PV = A/(1 +r)

For a sequence of non-uniform payments, we can work backward through the years as we did above.

4 0

3 years ago

Please help me solve for x

Masja [62]

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4 0

3 years ago

Whats the step by step and answer for 15h+30<10h-45

Andreyy89

First, subtract 10h from 15h so the variables are on the same side. You get:
5h+30<-45

Then, you can solve the inequality like any other equation. Subtract 30 from -45:
5h<-75

Finally, divide -75 by 5 to get your answer:
h<-15

7 0

3 years ago

Lin's father is paying for a $30 meal. He has a 15%-off coupon for the

Natali [406]

Answer:

$27.03

Step-by-step explanation:

Multiply: .15 times 30 =4.5 (to get from a percent to a decimal, move the decimal twice to the right.

Subtract: 30 minus 4.5= 25.5

Multiply: 25.5 times 1.06= 27.03

$27.03

Hope this helped!! :)

Brainliest?!?!

Stay safe and have a wonderful day/afternoon/night!!!

4 0

3 years ago