- #1
an_mui
- 47
- 0
If (log x)^2 + log x^3 - 27 = log x^4 - log x^7, solve for x to 2 decimal places
(log x)^2 + log x^5 - 27 = log x^4 - log x^7
(log x)^2 + 6 log x - 27 = 0
Let log x be a
a^2 + 6a - 27 = 0
(a + 9)(a - 3) = 0
a = 3 or - 9
log x = 3 or log x = -9
... x = 10^3 or 10^-9
Please help me because i am pretty sure this answer is wrong since the question asks us to solve x to 2 decimal places. however, i don't know what mistake i made so any help is appreciated! Thanks again!
(log x)^2 + log x^5 - 27 = log x^4 - log x^7
(log x)^2 + 6 log x - 27 = 0
Let log x be a
a^2 + 6a - 27 = 0
(a + 9)(a - 3) = 0
a = 3 or - 9
log x = 3 or log x = -9
... x = 10^3 or 10^-9
Please help me because i am pretty sure this answer is wrong since the question asks us to solve x to 2 decimal places. however, i don't know what mistake i made so any help is appreciated! Thanks again!