Exponentials and Logarithms? (see questions below)

The graph of #y=ab^x# passes through the points #(2,400)# and #(5,50)#.
a) find the values of the constants a and b
b) given that #ab^x < k#, for some constant #k > 0#, show that #x > log(1600/k)/log(2)# where log means log to any valid base

2 Answers
Mar 11, 2018

Shown below...

Explanation:

The equation #y = ab^x # is satisfied by #(2,400) # and # (5,50) #

Substituting we get..

#400 = ab^2 #

#50 = ab^5 #

Manipulating the first equation: #a = 400/b^2 #

Substituting into the second equation...

#50 = 400/b^2 * b^5 #

#=> 50 = 400b^3 #

#=> 50/400 = b^3 #

#=> b = root3(50/400) = root3 (1/8) = 1/2 #

Substituting into first equation...

#400 = a/4 => a = 1600#

#=> y = 1600 * (1/2)^x #

If #1600*(1/2)^x < k #

Take logs:

#log 1600*(1/2)^x < logk #

#=> log1600 + log(1/2)^x < logk #

#=> log1600 + x log(1/2) < logk #

#=> xlog(1/2) < logk - log1600 #

#=> xlog(1/2) < log(k/1600) #

#=> -xlog(2) < log(k/1600) #

#=> x > - log(k/1600)/ ( log(2) ) #

#=> x > log(1600/k) / log(2) #

Assuming you know your log laws!

#log(a) + log(b) = log(ab) #

#log(a)-log(b) = log(b/a) #

#klog a = loga^k => -loga = log(1/a) #

Mar 11, 2018

Part a

Given:

#y = ab^x#

Use the point #(2,400)# to write equation [1]:

#400 = ab^2" [1]"#

Use the point the point #(5,50)# to write equation [2]:

#50=ab^5" [2]"#

Divide equation [1] by equation [2]:

#400/50 = (ab^2)/(ab^5)#

Please observe that #a/a# becomes 1:

#8 = (cancel(a)b^2)/(cancel(a)b^5) = b^2/b^5#

We know that #b^(x_1)/b^(x_2) = b^(x_1-x_2)#

#8 = b^(2-5)#

#b^-3=8#

#b = 8^(-1/3)#

#b = 1/2#

Substitute the value of b into equation [1] to find the value of a:

#400 = a(1/2)^2#

#a = 1600#

Part b
Given:

#1600(1/2)^x < k#

Use the general logarithm of any base on both sides:

#log(1600(1/2)^x) < log(k)#

Use the property of logarithms that allows one to expand the product into a sum:

#log(1600)+ log((1/2)^x) < log(k)#

Multiply both sides by -1:

#-log(1600)- log((1/2)^x) > -log(k)#

Add #log(1600)# to both sides:

#-log((1/2)^x) >log(1600) -log(k)#

Use the property of logarithms that says that the difference of two logarithms is the same as division.

#-log((1/2)^x) >log(1600/k)#

The negative of a logarithm is the same as inverting the argument:

#log(2^x) >log(1600/k)#

Use the property of logarithms that allows one to bring the exponent outside as a coefficient:

#xlog(2) >log(1600/k)#

Divide both sides by #log(2)#:

#x >log(1600/k)/log(2)# Q.E.D.