logarithms (NEED help!!)

[maths_arghh234]

hey sorry if this is in the wrong boards, i'm not sure to be honest where logarithms fit ...

please help with these questions, i've tried them but i just get confused each time i try.

2. use the subtraction 3^x =u to solve the equation 3^2x-3^2+x+14=0
given your answer correct to three decimal places.

10. a) show that if x>0 log(base a) x^k =klog(base a) x
b) solve the equaltion
log(base 10 ) (x^2+48) = log(base 10) x +2log (base 10) 4

would really appreciate your help .. thanks

 

[daon]

Is there a typo in "3^2x-3^2+x+14=0"?

3^2 is 9 and you would be solving 3^(2x) + x = -5 or u^2 + ln(u)/ln(3)=-5

 

[maths_arghh234]

hey .. yea there was it's .. ^2x-3^(2+x)+14=0
2+x is the powere of that 3 ( sorry i wasen't clear)

 

[daon]

Then you need to solve u^2+9u+14=0. That shouldn't be hard for you, correct?

When you get your two answers, say u=a and u=b, then 3^x=a, 3^x=b, then solve for x in both.