What is x if #log(7x-12) - 2 log(x)= 1 #?

1 Answer
Oct 22, 2015

Answer:

Imaginary Roots

Explanation:

I think roots are imaginary
You may know that #log a^n = n log a#
So, #2 log x = log x^2#
Thus the equation becomes
#log (7x -12) - logx^2 = 1#
Also you may know
#log a - log c = log (a /c)#
Hence the equation reduces to
log #(7x - 12) / x^2 = 1#
You may also know,
if log a to base b is = c, then
#a = b^c#
For #log x# the base is 10
So the equation reduces to
#(7x - 12) / x^2 = 10^1 = 10#
or
#(7x - 12) = 10 * x^2 #
ie #10 * x^2 - 7x + 12 = 0#
This is a quadratic equation and the roots are imaginary, since #4 * 10 * 12 > 7^2#