log problem

[ruffnite]

Hi. I know I'm not doing this right, please point out where I went wrong. Thx.

Solve for x:
2^((x^2)-5x) = 8
log2^((x^2)-5x) = log8
(x^2-5x)log2 = log8
x^2-5x = (log8/log2)
x^2-5x = 3
x(x-5)=3

x = 3
x = 8

 

[garf]

Hi. I know I'm not doing this right, please point out where I went wrong. Thx.

Solve for x:
2^((x^2)-5x) = 8
log2^((x^2)-5x) = log8
(x^2-5x)log2 = log8
x^2-5x = (log8/log2)
x^2-5x = 3

UP TO HERE YOU ARE RIGHT. WHY DON'T YOU EQUAL TO 0 YOUR EQUATION, YOU WILL GET A SECOND GRADE EQUATION; IN THAT CASE YOU ALWAYS HAVE THE GENERAL FORMULA (HINT: YOU WON'T FIND INTEGER SOLUTIONS).
GARF

 

[mmm4444bot]

please point out where I went wrong.

Solve for x:

x^2 - 5x = 3

x(x - 5) = 3 Here is where you went wrong. We only factor when the other side is zero.

(Look up the Zero-Product Property.)

x^2 - 5x = 3 is a quadratic equation. Put it in General Form, and use the Quadratic Formula, to solve.

Also, you could have saved some work, if you had realized at the beginning that 8 is a power of 2, because there is no need to take the logarithm of both sides.

\(\displaystyle 2^{x^2 - 5x} = 8\)

\(\displaystyle 2^{x^2 - 5x} = 2^3\)

Since the base is the same on each side, the exponents must be equal.

\(\displaystyle x^2 - 5x = 3\)

 

[swaroop]

You are correct till your last step but you are not correct in solving the equation
x^2-5x-3=0
x=(5+-sqrt(25-4*-3))/2
x=(5+sqrt(37))/2 or (5-sqrt(37))2
x=5.24 or -0.54