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DerKrebs [107]
3 years ago
12

Let f and g be the functions defined by f(x) = 10^ (x+2 / 3) and g(x) = log (x3 / 100) for all positive real numbers,

Mathematics
1 answer:
Vaselesa [24]3 years ago
4 0

Answer:

F(x) and g(x) are not inverse functions.

Step-by-step explanation:

In order to calculate the inverse function of a function, we have to isolate X and after that, we change the variables.

As our function f(x) is a exponentian function, we can apply logarithm with base 10 (log) in both sides in order to isolate X. Remember that log10=1.

[tex]y=10^{(x+\frac{2}{3}) }\\\\log y=log 10^{(x+\frac{2}{3}) }\\log y = (x+\frac{2}{3}) . log10\\\frac{log Y}{log10} = (x+\frac{2}{3})\\\frac{log Y}{1} = (x+\frac{2}{3})\\log Y-\frac{2}{3}=x[/tex]

Now we change the variables.

F(x)=log x-\frac{2}{3}

F(x) and g(x) are not inverse functions.

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The simple interest on a certain sum of money for 2 years at 5% per annum is Rs 320. What will be the compound interest on the s
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Answer:

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Step-by-step explanation:

Find the <u>principal</u> amount invested.

<u>Simple Interest Formula</u>

I = Prt

where:

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  • t = time (in years)

Given:

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Substitute the given values into the formula and solve for P:

⇒ 320 = P(0.05)(2)

⇒ 320 = P(0.1)

⇒ P = 3200

<u>Compound Interest Formula</u>

\large \text{$ \sf I=P\left(1+\frac{r}{n}\right)^{nt} -P$}

where:

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  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

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Substitute the given values into the formula and solve for I:

\implies \sf I=3200\left(1+\frac{0.05}{1}\right)^{2} -3200

\implies \sf I=3200\left(1.05\right)^{2} -3200

\implies \sf I=3200\left(1.1025\right) -3200

\implies \sf I=3528-3200

\implies \sf I=328

Therefore, the compound interest on the same sum for the same time at the same rate is Rs 328.

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