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Masja [62]
2 years ago
13

how to solve for 8^x=2 (I know the answer is 1/3 but i have to show work and i don't know the process to solve it)

Mathematics
1 answer:
Lunna [17]2 years ago
5 0

Answer:

  x = 1/3 . . . . math facts or logarithms are involved; take your pick

Step-by-step explanation:

When you are solving for a variable that is in an exponent, logarithms are often useful. Taking the log of both sides of this equation, you have ...

  log(8^x) = log(2)

Using the rules of logarithms, that is ...

  x·log(8) = log(2)

  x = log(2)/log(8) . . . . . divide by the coefficient of x

You can find the value of this on your calculator, and it will tell you the value is  0.333333333333 or as many digits as your calculator displays. That is a clue that the exact answer is probably 1/3.

__

You should recognize that 8 = 2·2·2 = 2^3, so log(8) = 3log(2) and the above solution becomes ...

  x = log(2)/(3log(2)) = 1/3

__

Recognizing that 8 = 2^3, you can make that substitution into the original equation to get ...

  (2^3)^x = 2

  2^(3x) = 2^1

  3x = 1 . . . . . . . matching exponents; equivalent to taking logs base 2

  x = 1/3 . . . . . .  divide by 3

__

All of the above using 2 as a base of exponents is just dancing around the fact that you already know the math fact ...

  8^x = 2 = 8^(1/3)

  x = 1/3 . . . . . equating exponents; equivalent to taking logs base 8

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Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the in
RSB [31]

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

s^2+5s-2s-10=0

s(s+5)-2(s+5) =0

(s-2) (s+5) =0

s=2, s=-5

Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t)   ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10  = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

B=2

Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)

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