Question

help with logs!!!

1. e^2x-4e^x+3=0
2. 2log(base3)(x+4)=log(base3)9 + 2
3. log(base2)(x-6) + log(base2)(x-4)-log(base2)x=2
4. x^2x + 2^x - 12 = 0
5. 2 ^ 2x + 3^x - 2 = 0
6. e^2x - 2 e^x - 3 = 0

Answer 1

I) put y=e^x
then your eqn will be y^2-4y+3=0 use Quadratic equation to solve this

Answer 2

2log(base3)(x+4)=log(base3)9 + 2
Implies 2log(base3)(x+4)-log(base3)9=2
log(base3){[(x+4)^2]/9}=2
then exponentiate wit 3 and solve

Answer 3

same as above for 3

Answer 4

what do you mean by exponentiate?

Answer 5

if log(basek)m=n then k^n=m

Answer 6

so 3^2 = (x+4)^2/9?

Answer 7

for example in your qn
log(base3){[(x+4)^2]/9}=2
on exponentiation [(x+4)^2]/9=3^2

Answer 8

Yeah you r doing good

Answer 9

i have x^2 + 8 x - 65 = 0
is that right?

Answer 10

@Joseph91