How do you solve #log _ 2 (4x-8)=1#?
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"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
Jan 31, 2016
# x = 5/2 #
Explanation:
using :
# log_b a = n hArr a = b^n # then
# log_2 (4x - 8 ) = 1 → 4x - 8 = 2^1 = 2 # so 4x - 8 = 2
# → 4x = 10 → x = 10/4 = 5/2 #
