What happened to the "=0" part of the equation?Jade said:(3x-4)(2x+1)
x=4/3 x=-1/2
The rules aren't meant to be anything "deep". Just memorize them.Jade said:I think I am having trouble figuring out the log rules.
Assuming you mean "logb" to be "log<sub>b</sub>", these "rules" come from the basic definition of logarithms. Since b<sup>0</sup> = 1, then log<sub>b</sub>(1) = 0; since b<sup>1</sup> = b, then log<sub>b</sub>(b) = 1.Jade said:logb 1 = 0
logb b = 1
"Times" inside is "plus" outside; "divide" inside is "minus" outside; "to the power" inside is "times" outside. Just memorize 'em.Jade said:logb (xy) = logb x + logb y
logb (x/y) = logb x - logb y
logbx^r = r logb x
What you have posted means "1 + 2e<sup>5</sup>x = 15". Is that what you meant?Jade said:Same when e is involved.... for example I have 1+2e^5x=15
Do you mean "1 + 2e^(5x)" (that is, 1 + 2e<sup>5x</sup>)?Jade said:1 + 2e(^5x) = 15