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WILL MARK BRAINLIEST The amount (A) that principal (P) will be worth after t years at interest rate (r) compounded annually is

A = P(1 + r)

t


Suppose $6,000 is invested at 5.5% and yields a total of $7,255. How many years was it invested?

Respuesta :

MisterBrainly

Put in the formula

[tex]\\ \rm\Rrightarrow A=P(1+r)^t[/tex]

[tex]\\ \rm\Rrightarrow 7255=6000(1+0.055)^t[/tex]

[tex]\\ \rm\Rrightarrow 1.20917=(1.055)^t[/tex]

[tex]\\ \rm\Rrightarrow ln1.20917=tln1.055[/tex]

[tex]\\ \rm\Rrightarrow t=\dfrac{ln1.20917}{ln1.055}[/tex]

[tex]\\ \rm\Rrightarrow t=3.5years[/tex]

semsee45

Answer:

4 years (nearest year) or

3.5 years (nearest tenth) or

3 years, 6 months and 17 days (most accurate)

Step-by-step explanation:

Annual Compound Interest Formula

[tex]\large \text{$ \sf A=P\left(1+r\right)^{t} $}[/tex]

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • t = time (in years)

Given:

  • A = $7,255
  • P = $6,000
  • r = 5.5% = 0.055

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

[tex]\implies \sf 7255=6000(1+0.055)^t[/tex]

[tex]\implies \sf 7255=6000(1.055)^t[/tex]

[tex]\implies \sf \dfrac{7255}{6000}=(1.055)^t[/tex]

Take natural logs of both sides:

[tex]\implies \sf \ln \left(\dfrac{7255}{6000}\right)= \ln (1.055)^t[/tex]

[tex]\textsf{Apply the power law}: \quad \ln x^n=n \ln x[/tex]

[tex]\implies \sf \ln \left(\dfrac{7255}{6000}\right)= t\ln (1.055)[/tex]

[tex]\implies \sf t=\dfrac{\ln \left(\dfrac{7255}{6000}\right)}{\ln (1.055)}[/tex]

[tex]\sf \implies t=3.547416819...[/tex]

Therefore, the number of years the principal was invested is:

  • 4 years (nearest year)
  • 3.5 years (nearest tenth)
  • 3 years, 6 months and 17 days

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