Ok. so I know I'm dumb. But can someone help me solve this problem?
log3(x)= log3(2)-log3(x-2)
solve for x
the 3s are supposed to be subscripts obviously
thanks!!!
Ok. so I know I'm dumb. But can someone help me solve this problem?
log3(x)= log3(2)-log3(x-2)
solve for x
the 3s are supposed to be subscripts obviously
thanks!!!
The right side can be simplified to log3(2/(x-2)), and then you can drop the logs all together and be left with x=2/(x-2).
try a laxative
kele is right. equal bases, equal exponents. logs are an inverse exponential expression.
I dont think you can just take the logs away
really?
please explain in more detail. I don't have any laxatives on me.
Sure you can. You have log3(x)=log3(2/(x-2)), so you're just operating on both sides of the same equation in the same exact way with the log3. Raise 3 to each side of the equation ( 3^log3(x)=3^log3(2/(x-2)) ), and the 3's and log3's drop out leaving you with the simplification that I gave above.
Please for the love of god I am going to curl up on the floor and howl if someone does not explain this to me in detailed, step by step instructions.
thank you you are my holy savior, I will love you until my dying day
If LHS=RHS => 3^(LHS)=3^(RHS)
Use: 3^(a-b)=3^a/3^b
=>3^(RHS)=3^((log3(2)-log3(x-2))=3^log3(2)/3^log3(x-2)=2/x-2
Also, 3^(LHS)=x
=>x(x-2)=2
Okay here is a challenging one: Find a formula for the inverse of f(x)=cos(2x) for 0 <= x<= pi/2
inverse = g(x) = arccos(x)/2 ?
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