SOLUTION: Could someone please help me with this problem? Solve. Where appropriate,include approximations to the nearest thousandth. if no solution exists, state this. log (x-3) + log

Algebra.Com
Question 341711: Could someone please help me with this problem?
Solve. Where appropriate,include approximations to the nearest thousandth. if no solution exists, state this.
log (x-3) + log x= 1
thank you for your help.

Answer by user_dude2008(1862)   (Show Source): You can put this solution on YOUR website!
log (x-3) + log x= 1


log [x(x-3)]= 1

x(x-3)=10^1

x^2-3x=10

x^2-3x-10=0

(x-5)(x+2)=0

x=5 or x=-2 (ignore)


Only solution is x=5

RELATED QUESTIONS

Can someone help me solve this problem? Where appropiate, include approximations to the... (answered by Fombitz)
Solve. Where appropriate, include approximations to the nearest thousandth. If no... (answered by user_dude2008)
Please help I am really confused and could use a step by step break down. Solve. Where... (answered by ankor@dixie-net.com)
Need Help Please! I am failing this class.. 5. Solve. Where appropriate, include... (answered by rapaljer,josmiceli)
Solve. Where appropriate, include approximations to the nearest thousandth. If no... (answered by jsmallt9)
Solve. Where appropriate, include approximations to the nearest ten-thousandth. Ln... (answered by Fombitz)
Solve. Where appropriate, include approximations to the nearest thousandth. If no... (answered by edjones)
Solve. Where appropriate, include approximations to the nearest thousandth. If no... (answered by edjones)
solve. where appropriate, include approximations to the nearest thousandth. if no... (answered by solver91311)