Logarithm and Exponent Help?

Question

I have a bunch of questions to solve and am badly stuck at these three questions. I tried to solve them but don't seem to be going right. Besides I don't have the answers. Help is greatly appreciated.

Q1. An Earthquake measures 6.5; Aftershock measures 4.5. How many times bigger was the earthquake than the aftershock?

Q2. If a substance decays at a continuous rate of 7% per day and you need at least 120g of the substance in a week's time, how much must you have today?

Q3. 8^(3x-21) = 11^(2x+7)

Answer 1

1.)
(6.5)/(4.5) = (65/45) = (13/9) = about 1 4/9 times larger.

------------------------

2.)
A = Ao * e^(-rt)
120 = Ao * e^(-.07 * 7)
120 = Ao * e^(.49)
Ao = 120/(e^(-.49))
Ao = about 195.8779 or about 196 grams

-------------------------

3.)
8^(3x - 21) = 11^(2x + 7)
(3x - 21)ln(8) = (2x + 7)ln(11)
3xln8 - 21ln8 = 2xln11 + 7ln11
3xln8 - 2xln11 = 21ln8 + 7ln11
x(3ln8 - 2ln11) = 21ln8 + 7ln11
x = (21ln(8) + 7ln(11))/(3ln(8) - 2ln(11))

x = (21ln(8) + 7ln(11))/(3ln(8) - 2ln(11))
or
x = 7((3ln(8) + ln(11))/(3ln(8) - 2ln(11)))

Answer 2

well, I know that on the normal scale for earthquakes one point increase means the earthquake is 30% stronger.... so, the Earthquake would be 60% bigger than the aftershock.

for question three.

you multiply both sides by the log of (subscript) 8

so,

[log (subscript) 8]8^(3x-21) = [log (subscript) 8] 11^(2x+7)

log (subscript) 8 times 8 is 1

so

3x-21 = [log (subscript) 8] 11^(2x+7)

then distribute the right hand side of the equation to...

3x-21 = [2log(sub.)8]11x + [7log(sub.)8]11

isolate the variable

3x - [2log(sub.)8]11x = [7log(sub.)8]11 + 21

take x out of the left side of the equation to isolate the variable further

x{3-[2log(sub.)8]11} = [7log(sub.)8]11] + 21

divide both sides by the expression in the " { } "

and you get {3-[2log(sub.)8]11} over {3-[2log(sub.)8]11} which cancels itself out

and [7log(sub.)8]11] + 21 over 3-[2log(sub.)8]11

then you convert the logarithms into natural logs to rationalize the answer

and (ln = natural log)

7 ln 11 / ln 8 + 21 over 3 - 2 ln 11 / ln 8

common denominator is ln 8
add ln 8 / ln 8 to the 21 and 3 and you get

7 ln 11 / ln 8 + 21 ln 8 / ln 8 over 3 ln 8 / ln 8 - 2 ln 11 / ln 8

simplify and you get

7 ln 11 + 21 ln 8 over 3 ln 8 - 2 ln 11 = x

there's your answer

hope that helps.
email me for questions: [e-mail]

Answer 3

4.5*(what?)=6.5

so 6.5/4.5 = 65/45 = 13/9 This is what we were looking for (the multiplication factor)

Now,

we have a mass which we don't know... after one day the mass is 7% smaller = Mass (original) - .07(Mass orignal) = Mass(day 2) = Mass(orginal) * (1 - .07) = .93*Mass (orignal)

so on day three we come back to the mass and it is again 7% smaller
Mass (one day old) - .07*Mass (one day old) = Mass(2 days old) = Mass(orginal) * (1 - .07) = .93*Mass(one day old)

but we know that Mass (one day old) = .93*Mass (orginal) so of the orginal we have.... Mass (2 days old) = .93*.93*Mass(orginal)

Now we look at this and say, "hey it just looks like a multiply the orginal mass times .93^(days old) and i will have the amount of mass i will have on that day.

so 120grams = .93^7 Mass (orginal)
==> Mass (orginal) = 120 grams/ .93^7

For the next one you use logs

(3x-21)*log8 = (2x +7)*log11

so just use your calc or you can reduce it a bit more

3x-21 = (log11/log8)*2x + 7*(log11/log8)

x(3 - 2*(log11/log8)) = 21 + 7*(log11/log8)

x = (21 + 7*(log11/log8)) / ((3 - 2*(log11/log8))

Answer 4

at the same time as the bottom of a logarithm contained in the exponent and the bottom of an exponent are the same, the outcome's the argument of the logarithm: a^log?( b ) = b e^log?( b ) = b e^ln( b ) = b e^ln( 5 ) = 5

Answer 5

1.6.5/4.5 times

2.x-7*0.07x=120
x-0.47x=120
0.53x=120
x=120/0.53

3.(3x-10)8ln=(2x+7)ln11
(3x-10)/(2x+7)=ln8/ln11=>log8/log11
=0.9030/1.0414
=0.8671
now cross multiply and solve