Calculus Solving logarithmic equations

[an_mui]

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!

 

[amcavoy]

Simplify everything except for the first term and the constant to end up with something the resembles a quadratic equation. Instead of x, you will have something in terms of log(x). Now, use the quadratic formula to solve for log(x), and exponentiate.

 

[an_mui]

Sorry i thought I did what you said above. I simplified everything except for the first term and the constant. However, I ended up with log x = 3 or log x = -9.

 

[amcavoy]

Oh, yes, sorry I was confused when you wrote log(x⁵). Yeah, what you did looks good; I wouldn't worry about the 2 decimal places.

 

[HallsofIvy]

(log x)^2 + log x^5 - 27 = log x^4 - log x^7
(log x)^2 + 6 log x - 27 = 0

This is incorrect: (log x)^2+ 5 log x- 27= 4 log x- 7 log x= -3 log x so
(log x)^2+ 8 log x- 27= 0

Your quadratic equation is a^2+ 8a- 27= 0 which does not have rational solutions.

 

[amcavoy]

(log x)^2 + log x^5 - 27 = log x^4 - log x^7
(log x)^2 + 6 log x - 27 = 0
This is incorrect: (log x)^2+ 5 log x- 27= 4 log x- 7 log x= -3 log x so
(log x)^2+ 8 log x- 27= 0
Your quadratic equation is a^2+ 8a- 27= 0 which does not have rational solutions.

I'm confused about your problem here. In your first line (of the OP), you said the second term was log(x³). The next line you say it is log(x⁵). Your solution is correct assuming it is x cubed. However, if it is x to the fifth you need to follow Halls' suggestion.