# Calculus Solving logarithmic equations

1. Nov 28, 2005

### 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!

2. Nov 28, 2005

### 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.

3. Nov 28, 2005

### 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.

4. Nov 28, 2005

### 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.

5. Nov 29, 2005

### HallsofIvy

Staff Emeritus
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.

6. Nov 29, 2005

### amcavoy

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.