What is the solution set of #log_7 (x+3) + log_ 7 (x-5) = 1#?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Alan N.
Mar 22, 2018

#x = 1+sqrt23 approx 5.7958#

Explanation:

#log_7 (x+3) + log_7 (x-5) =1#

Applying #log_n a + log_n b = log_n ab#

#log_7 (x+3)(x-5) = 1#

Now, # 1 = log_7 7#

#:. log_7 (x+3)(x-5) = log_7 7#

If #log_n a = log_n b# then #a=b#

Thus, #(x+3)(x-5) =7#

#x^2 -2x-15 =7#

#x^2-2x-22 =0#

Apply quadratic formula

#x =(2+-sqrt(4+4xx22))/2#

#= 1+- sqrt92/2 =1+- (2sqrt23)/2#

#= 1+-sqrt23#

Remember that #log_n (x)# is undefined for #x<0#

Then, since #sqrt23>1#, we can reject the negative result.

NB: since #1+sqrt23>5#, we can retain the positive result

#:. x=1+sqrt23 approx 5.7958#

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