Question 735204
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Please show how to solve:

Log base10(n^2 � 90n) = 3

Thank you
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To this author, what the other person who responded wrote, as a solution to this equation, doesn't make any sense.
That's, "....n^2 - 90n = log base 10 (3) = .48
use quadratic formula and we get (note that b is --90 or 90)
n=(90+square root (90^2-4*1*-.48)) / 2 = 90"

{{{log (10, (n^2 - 90n)) = 3}}} <== Base 10 is OPTIONAL, because we work in the decimal system, so usually, base 10 is NOT entered.
Since the log argument {{{n^2 - 90n}}} MUST be greater than 0, we have: {{{n^2 - 90n > 0}}} ===> {{{n(n - 90) > 0}}}.
The SOLUTIONS to the INEQUALITY, 0 and 90 are the CRITICAL POINTS, with 3 intervals: Interval 1: n-values < 0
                                                                                                                                                           Interval 2: 0 < n-values < 90
                                                                                                                                                           Interval 3: n-values > 90

When tested, we find that the n-values that'll satisfy the INEQUALITY are < 0, and > 90. So, based on that, we get:
 {{{log (10, (n^2 - 90n)) = 3}}}, with n < 0, or > 90.
               {{{n^2 - 90n = 10^3}}} ---- Converting to EXPONENTIAL form
               {{{n^2 - 90n = "1,000"}}}
 {{{n^2 - 90n - "1,000" = 0}}}
(n - 100)(n + 10) = 0
                 n - 100 = 0           OR          n + 10 = 0 ----- Setting each factor equal to 0
                           n = 100       OR                   n = - 10

As seen, 100 is > 90, and - 10 is < 0, so both solutions are VALID/ACCEPTABLE!</font></font></font></b></pre>