Asked by Julia
If log(2a+3b/2)=1/2log(2a)+1/2log(3b), where a> 0,b >0, show that 4a^2 + 9b^2 = 12ab Show all calculations
Answers
Answered by
Reiny
Unless otherwise stated we must assume your base of the log is 10
I will also assume you mean:
log((2a+3b)/2)=1/2log(2a)+1/2log(3b)
log((2a+3b)/2)=1/2(log(2a)+log(3b))
log((2a+3b)/2)= (log(6ab)^(1/2))
(2a+3b)/2 = √(6ab)
square both sides:
(4a^2 + 12ab + 9b^2)/4 = 6ab
4a^2 + 12ab + 9b^2 = 24ab
4a^2 + 9b^2 = 12ab , as required
I will also assume you mean:
log((2a+3b)/2)=1/2log(2a)+1/2log(3b)
log((2a+3b)/2)=1/2(log(2a)+log(3b))
log((2a+3b)/2)= (log(6ab)^(1/2))
(2a+3b)/2 = √(6ab)
square both sides:
(4a^2 + 12ab + 9b^2)/4 = 6ab
4a^2 + 12ab + 9b^2 = 24ab
4a^2 + 9b^2 = 12ab , as required
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