Express in terms of???plz help?

Question

If logª x = p, express logª 1/x^2 in terms of p. (the a is actually at the bottom of g..not a power)

Also, the gradient of a point (x,y) is given by dy/dx = e^(-2x). If the curve passes through (0,2), find y in terms of x.

Answer 1

logª 1/x^2 = log(1) - log(x^2)

= 0 - 2log(x)

= -2p

---

dy/dx = e^(-2x)

y = e^(-2x)/(-2) + C

Substitute (0,2) to find C

2 = e^(0)/(-2) + C

C = 2 + 1/2 = 2.5

y = -0.5*e^(-2x) + 2.5

Answer 2

a.) log (base a) x = p ----> a^p = x
log (base a) 1/x = log (base a) 1 - log (base a) x = 0 - p = -p.

Recall that 1/(x)² = (1/x)² = (1/x)(1/x). Then this is the case:

log (base a) 1/(x)² = log (base a) (1/x)² = log (base a) (1/x)(1/x) = log (base a) (1/x) + log (base a) (1/x) = -p + (-p) = -2p.

So log (base a) 1/(x)² = -2p.

b.) dy/dx = e^(-2x) ----> dy = e^(-2x) dx. This implies:

y = ƒ e^(-2x) dx = -½ ƒ e^(-2x)(-2) dx = -½ e^(-2x) + c, where ƒ represents the integral symbol and c is a constant of integration.

Given (0,2) is a point on the graph of the above equation, y = -½ e^(-2x) + c, if we let x = 0 and y = 2, then this is true:

2 = -½ e^-2(0) + c
2 = -½ e^(0) + c
2 = -½ (1) + c
2 = -½ + c
2 + ½ = c
5/2 = c

So our complete equation for y in terms of x is:

y = -½ e^(-2x) + 5/2

Answer 3

Interesting, no one answered the gradient question.

if you know that dy/dx=e^(-2x) then multiply both sides by dx and integrate and you get that y=(-1/2)(e^(-2x))+C and since it must pass through (0,2), plug in the point and you get that C=2.5

So the function y is
y=(-1/2)(e^(-2x))+2.5

Answer 4

If logª x = p, express logª 1/x^2 in terms of p

logª 1/x^2 = -2p

Answer 5

If log (base a) x = p then a ^ p = x.
So, log (base a) (1/x ^ 2) = log (base a) (1/((a ^ p) ^ 2)) = -2p

Answer 6

Given [log(x)]^2 = p

We manipulate this expression using log identities.
[log(1/x^2)]^2

[log(x^(-2))]^2

[(-2)log(x)]^2

(-2)^2 [log(x)]^2
4 [log(x)]^2

But [log(x)]^2 = p, so our final answer is

4p