Question

log4 (2x+3) - 2log4 x= 2

4 is base

Answer 1

answer is 9

Answer 2

what i dont understand

Answer 3

log_b(A) - log_b(B) = log_b(A/B)

log_b(A) + log_b(B) = log_b(AB)

log_b(A^B) = B*log_b(A)

Answer 4

why nine 9

Answer 5

\[\log_4(2x + 3) - \log(x^2) = 2\]\[\log_4\left( \frac{ 2x+3 }{ x^2 } \right) = 2\]\[\frac{ 2x+3 }{ x^2 } = 4^2\]\[16x^2 - 2x - 3 = 0\]solve for x

Answer 6

he's lying. he thinks he's funny

Answer 7

ah ok .. thank you

Answer 8

i think i will use quadratic equation

Answer 9

@Euler271

Answer 10

good idea. i would too

Answer 11

can i use quadratic equation ?

Answer 12

with a second order polynomial: always

Answer 13

where did you get the 16x^2