What is the effect of error on a calculation, and what are the components of error that are often considered?

Step 1: Apply the logarithm property ln a + ln b = ln(ab) for a>0, b>0. ln[(x+1)(x−1)] = ln 3 Step 2: Since ln is one-to-one, equate the arguments. (x+1)(x−1) = 3 Step 3: Expand the left side. x² − 1 = 3 Step 4: Solve for x². x² = 4; x = ± 2 Step 5: Determine the domain where both arguments are positive: x+1 > 0 and x−1 > 0. x > 1 Step 6: Check candidate solutions in the domain. For x=2 > 1: valid. Verify: ln(3) + ln(1) = ln 3 + 0 = ln 3. For x=−2 < 1: ln(−1) undefined in reals. Invalid. Final answer x=2

What is the effect of error on a calculation, and what are the components of error that are often considered?

Step 1: Apply the logarithm property ln a + ln b = ln(ab) for a>0, b>0. ln[(x+1)(x−1)] = ln 3 Step 2: Since ln is one-to-one, equate the arguments. (x+1)(x−1) = 3 Step 3: Expand the left side. x² − 1 = 3 Step 4: Solve for x². x² = 4; x = ± 2 Step 5: Determine the domain where both arguments are positive: x+1 > 0 and x−1 > 0. x > 1 Step 6: Check candidate solutions in the domain. For x=2 > 1: valid. Verify: ln(3) + ln(1) = ln 3 + 0 = ln 3. For x=−2 < 1: ln(−1) undefined in reals. Invalid. Final answer x=2

Solved: What is the effect of error on a calculation, and…

Apply the logarithm property $\ln a + \ln b = \ln(ab)$ for $a>0$ , $b>0$ .
$\ln[(x+1)(x-1)] = \ln 3$
Since $\ln$ is one-to-one, equate the arguments.
$(x+1)(x-1) = 3$
Expand the left side.
$x^2 - 1 = 3$
Solve for $x^2$ .
$x^2 = 4$ $x = \pm 2$
Determine the domain where both arguments are positive
$x+1 > 0$ and $x-1 > 0$ .
$x > 1$
Check candidate solutions in the domain.
For $x=2 > 1$ : valid.
Verify: $\ln(3) + \ln(1) = \ln 3 + 0 = \ln 3$ .

For $x=-2 < 1$ : $\ln(-1)$ undefined in reals. Invalid.

Final answer

x=2

What is the effect of error on a calculation, and what are the components of error that are often considered?

Answer

Step-by-step solution

Apply the logarithm property $\ln a + \ln b = \ln(ab)$ for $a>0$ , $b>0$ .

Since $\ln$ is one-to-one, equate the arguments.

Expand the left side.

Solve for $x^2$ .

Determine the domain where both arguments are positive

Check candidate solutions in the domain.

Try one yourself

Still stuck on this one?

More Mathematics Questions

What is the effect of error on a calculation, and what are the components of error that are often considered?

Answer

Step-by-step solution

Apply the logarithm property ln⁡a+ln⁡b=ln⁡(ab)\ln a + \ln b = \ln(ab)lna+lnb=ln(ab) for a>0a>0a>0, b>0b>0b>0.

Since ln⁡\lnln is one-to-one, equate the arguments.

Expand the left side.

Solve for x2x^2x2.

Determine the domain where both arguments are positive

Check candidate solutions in the domain.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Apply the logarithm property $\ln a + \ln b = \ln(ab)$ for $a>0$ , $b>0$ .

Since $\ln$ is one-to-one, equate the arguments.

Solve for $x^2$ .