Linearize the following equations using natural logarithms: P = P0 (1 - kh)ⁿ, I = B e^-lambda t, and E = ATⁿ…

Question

Linearize the following equations using natural logarithms: P = P0 (1 - kh)ⁿ, I = B e^-lambda t, and E = ATⁿ. Identify the slope and intercept in the y=mx+c form.

ScanSolve AI Tutor · Accepted Answer

Here are the linearized forms of the equations from the paper:

1. For the equation $P = P_0 (1 - kh)^n$ ()

Take the natural logarithm ( $\ln$ ) of both sides of the equation.
$\ln(P) = \ln(P_0 (1 - kh)^n)$
Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .
$\ln(P) = \ln(P_0) + \ln((1 - kh)^n)$ $\ln(P) = \ln(P_0) + n \ln(1 - kh)$
Rearrange into the linear form $y = mx + c$ .
Let $y = \ln(P)$ and $x = \ln(1 - kh)$ .
The linearized equation is:
$\boxed{\ln(P) = n \ln(1 - kh) + \ln(P_0)}$
Here, the slope $m_{slope} = n$ and the y-intercept $c_{intercept} = \ln(P_0)$ .

2. For the equation $I = B e^{-\lambda t}$ ()
Take the natural logarithm ( $\ln$ ) of both sides of the equation.
$\ln(I) = \ln(B e^{-\lambda t})$
Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .
$\ln(I) = \ln(B) + \ln(e^{-\lambda t})$ $\ln(I) = \ln(B) - \lambda t$
Rearrange into the linear form $y = mx + c$ .
Let $y = \ln(I)$ and $x = t$ .
The linearized equation is:
$\boxed{\ln(I) = (-\lambda)t + \ln(B)}$
Here, the slope $m_{slope} = -\lambda$ and the y-intercept $c_{intercept} = \ln(B)$ .

3. For the equation $E = AT^n$ ()
Take the natural logarithm ( $\ln$ ) of both sides of the equation.
$\ln(E) = \ln(AT^n)$
Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .
$\ln(E) = \ln(A) + \ln(T^n)$ $\ln(E) = \ln(A) + n \ln(T)$
Rearrange into the linear form $y = mx + c$ .
Let $y = \ln(E)$ and $x = \ln(T)$ .
The linearized equation is:
$\boxed{\ln(E) = n \ln(T) + \ln(A)}$
Here, the slope $m_{slope} = n$ and the y-intercept $c_{intercept} = \ln(A)$ .

4. For the equation $I = I_0 \exp\left(\frac{eV}{kT}\right)$ ()
Take the natural logarithm ( $\ln$ ) of both sides of the equation.
$\ln(I) = \ln\left(I_0 \exp\left(\frac{eV}{kT}\right)\right)$
Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .
$\ln(I) = \ln(I_0) + \ln\left(\exp\left(\frac{eV}{kT}\right)\right)$ $\ln(I) = \ln(I_0) + \frac{eV}{kT}$
Rearrange into the linear form $y = mx + c$ .
Let $y = \ln(I)$ and $x = V$ .
The linearized equation is:
$\boxed{\ln(I) = \left(\frac{e}{kT}\right)V + \ln(I_0)}$
Here, the slope $m_{slope} = \frac{e}{kT}$ and the y-intercept $c_{intercept} = \ln(I_0)$ .

Linearize the following equations using natural logarithms: P = P0 (1 - kh)ⁿ, I = B e^-lambda t, and E = ATⁿ. Identify the slope and intercept in the y=mx+c form.

Step-by-step solution

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .

Rearrange into the linear form $y = mx + c$ .

Still stuck on this one?

More Mathematics Questions

Linearize the following equations using natural logarithms: P = P0 (1 - kh)ⁿ, I = B e^-lambda t, and E = ATⁿ. Identify the slope and intercept in the y=mx+c form.

Step-by-step solution

Take the natural logarithm (ln⁡\lnln) of both sides of the equation.

Use logarithm properties ln⁡(AB)=ln⁡(A)+ln⁡(B)\ln(AB) = \ln(A) + \ln(B)ln(AB)=ln(A)+ln(B) and ln⁡(Xn)=nln⁡(X)\ln(X^n) = n \ln(X)ln(Xn)=nln(X).

Rearrange into the linear form y=mx+cy = mx + cy=mx+c.

Take the natural logarithm (ln⁡\lnln) of both sides of the equation.

Use logarithm properties ln⁡(AB)=ln⁡(A)+ln⁡(B)\ln(AB) = \ln(A) + \ln(B)ln(AB)=ln(A)+ln(B) and ln⁡(ex)=x\ln(e^x) = xln(ex)=x.

Rearrange into the linear form y=mx+cy = mx + cy=mx+c.

Take the natural logarithm (ln⁡\lnln) of both sides of the equation.

Use logarithm properties ln⁡(AB)=ln⁡(A)+ln⁡(B)\ln(AB) = \ln(A) + \ln(B)ln(AB)=ln(A)+ln(B) and ln⁡(Xn)=nln⁡(X)\ln(X^n) = n \ln(X)ln(Xn)=nln(X).

Rearrange into the linear form y=mx+cy = mx + cy=mx+c.

Take the natural logarithm (ln⁡\lnln) of both sides of the equation.

Use logarithm properties ln⁡(AB)=ln⁡(A)+ln⁡(B)\ln(AB) = \ln(A) + \ln(B)ln(AB)=ln(A)+ln(B) and ln⁡(ex)=x\ln(e^x) = xln(ex)=x.

Rearrange into the linear form y=mx+cy = mx + cy=mx+c.

Still stuck on this one?

More Mathematics Questions

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(X^n) = n \ln(X)$ .

Rearrange into the linear form $y = mx + c$ .

Take the natural logarithm ( $\ln$ ) of both sides of the equation.

Use logarithm properties $\ln(AB) = \ln(A) + \ln(B)$ and $\ln(e^x) = x$ .

Rearrange into the linear form $y = mx + c$ .