How to solve Exponential eqations?

Question

here is an example

it is found that the intensity of light decreases as it passes through water. The intensity I units at a depth of x metres from the surface is given by I=Ia (10)^-kx, x≥ 0, where Ia units is the intensity at the surface. Based on the recordings taken by a diving team it was found that I=0.2Ia at a depth of 50 metres.

a) Find the value of k
b) Find the percentage of light remaining at 20 metres
c) find the depth at which the intensity would be half of that at the surface.

Answer 1

The equation is :
I = Ia 10^-kx

a) if x =50 then I = 0.2 Ia
So 0.2 = 10^-k50
taking log decimal :
log(0.2) = -k50 so k = -log0.2 /50 = (log5) /50 = 0.0140 m^-1

b) if x= 20m
I = Ia 10^-kx = Ia 10^-0.0140*20 = 0.52 Ia
So the percentage is 52%

c) If the intensity is the half I = 0.5 Ia
10^-kx = 0.5
So x = log 2 /k = 21.5 m

Answer 2

Question a)

let I = Ia (10) ^-kx
rewrite I/Ia = (10) -kx
inverse the fraction Ia/I = (10) kx aand take the log
log (Ia/I) = kx where you find k = log (Ia/I) /x
here k = log (Ia/ 0.2 Ia)/50 = 0.01397

and I = Ia (10) ^-0.01397 x equation (1)

question b)

replace x by 20 in equation 1
I = Ia (10)^(-0.01397*20) =-0.525 Ia

question c)

At the depth x I = 0.5 Ia

Ia = 0.5 Ia (10)^(-0.01397*xh) where xh is the depth

2 = (10)^(-0.01397*xh)

and xh =(log 2) /0.01397) = 21.55m

hoping it helped

Answer 3

Hi

Let a = diving depth M , y = intensity at surface L = intensity at x metres, k a constant.

Then

L = y x 10 ^ -(ak) = 0.2y

divide both sides by y

a) 10 ^ (-0.7) = 0.2 approximately

Hence k = (0.7) x (0.02) = 0.014

b) 0.525 to 3. dp = 52.5%

c)I found the value to be 21.5M

Thankyou

Answer 4

72

Answer 5

light = surf light * 10^(-kx) ( x greater than zero ... measures depth

.2 = 10(-k 50)
take the log
log(.2) = 50 k
k = log(.2) / 50

A)>>>>k = -0.01397


B)>>>>? = (1)10^-(20*-.01397)

.2794 or roughly 28%

C) >>>.5 = 10^-(.01397 * x )
log (.5) = -.01397 x
[log(.5) / -.01397] = x
....
x = 21.54 m

Answer 6

L=(10)^(-kx)

LOGL = LOG(10)^(-kx) = (-kx)LOG(10) = -kx(1)

LOGL = -kx

LOG(.02) = -k(50)
-.7 = -k(50)

k = .014

L=(10)^((-.014)(50)) = .2

.5=(10)^((-.014)(x))
LOG.5 = LOG(10)^((-.014)(x))=-.014x
-.3=-.014x
x=21.4