Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x) I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday
Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x) I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday
Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x) I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday
I have done but I end up with log(base a)x^10 - log(base a)(x^6/16). It's he fraction I'm struggling with. I tried to make it not a fraction like you'd do with 1/x^2 becoming x^-2 but I'm not sure how to do it x on top.
Why has the sign changed in the second step from "-" to a "+"
Log rule - the minus 2 has been put as a power for (x^3/4). You have you interpret minuses as plussing a minus... If that makes sense. Basically, revise log rules so you're sure.
Ah it's so simple thank you! I didn't realise if you had a coefficient infront of the x that you had to take the 4th root of 16 to make it 4loga(2x) thought it was just left as 16