Question

given that log _{a}(3x-4a)+log _{a}3x=(2div log _{2}a)+log _{a}(1-2), where 0 14 years ago

Answer 1

What's this last bit? log _{a}(1-2)

Answer 2

wrong question....
actually is

given that \[\log _{a}(3x-4a)+\log _{a}3x=(2\div \log _{2}a)+\log _{a}(1-2a)\], where \[0<a<1\div2\], find x

Answer 3

i try half way then stuck at here...
\[\log _{a}(3x-4a)+\log _{a}3x=(2\div \log _{2}a)+\log _{a}(1-2a)\]\[\log _{a}(3x-4a)(3x)=(2\log _{2}2\div \log _{2}a)+\log _{a}(1-2a)\]\[\log _{a}(3x-4a)(3x)=(\log _{2}2^2\div \log _{2}a)+\log _{a}(1-2a)\]\[\log _{a}(3x-4a)(3x)=(\log _{2}4\div \log _{2}a)+\log _{a}(1-2a)\]\[\log _{a}(3x-4a)(3x)=\log _{a}4+\log _{a}(1-2a)\]\[\log _{a}(3x-4a)(3x)=\log _{a}(4)(1-2a)\]\[(3x-4a)(3x)=4(1-2a)\]\[9x^2-12ax=4-8a\]\[9x^2-12ax-4+8a=0\]\[9x^2+(-12a)x+(-4+8a)=0\]\[x=(-(-12a)\pm \sqrt{(-12a)^2-4(9)(-4+8a)})\div 2(9)\]\[x=(12a \pm \sqrt{144a^2-288a+144})\div 18\]\[x=(12a \pm 12\sqrt{a^2-2a+1})\div 18\]\[x=(12a \pm 12\sqrt{(a-1)^2})\div 18\]\[x=(12a \pm 12(a-1))\div 18\]\[x=(2a \pm 2(a-1))\div 3\]

Answer 4

Well I assume u have done all that correctly.

Now u have a range for a -> 0<a<1/2

So sub at those values to get a range for x.