HELP ! - Logs (1 Viewer)

[ghostman_on_3rd]

Solve for x.

Log(base a) 16x = 3log(base a) x



Thanks in advance

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Is x =0 and x = 4 and x = -4?

I don't have the real answer for this question

 

[lyounamu]

Solve for x.

Log(base a) 16x = 3log(base a) x



Thanks in advance

Log (basea) 16x = log (basea)x^3
x^3 = 16x
x^3- 16x = 0
x(x^2-16) = 0
x(x-4)(x+4) = 0
x = 0 , -4 or 4

BUT x > 0, so x = 4

 

[Drongoski]

Solve for x.

Log(base a) 16x = 3log(base a) x



Thanks in advance

log_a 16x = 3 log_a = log_a x³

Therefore x³ = 16 x

i.e. x(x - 4)(x + 4) = 0

Therefore x = 0, -4, 4

But as lyounamu pointed out (and I overlooked) the first 2 values are inadmissible.

 

[ghostman_on_3rd]

Thanks guys !