Question

y = 16log2x
find x

Answer 1

hmm is it y = 16^(log(2x)) or is that 2 base of log?

Answer 2

I don't know, that's how the exercise showed it...

Answer 3

hmm i dont think it's base then..

Answer 4

let me pull up a list of log properties

Answer 5

Thx

Answer 6

take log on both sides:
log(y) = log(16)^(log(2x))
log(a)^b = b*log(a), so
log(y) = log(2x)*log(16)

Answer 7

log(2x) = log(y)/log(16)
10^log(2x) = 10^(log(y)*log(16))
2x = (10^(log(y)))^log(16) = y^log(16)
x = y^(log(16))/2

Answer 8

I don't get this. Can you help me @ash2326 ?

Answer 9

Options are:
a) 2y
b) sqrt y
c) 4 sqrt y
d) 4y

Answer 10

We have
\[y= 16^{log_{2}x}\]
we know that 16=2^4
so
\[y= (2^4)^{log_{2}x}\]
so we get
\[y= 2^{4log_{2}x}\]
or
we can write
\[y= 2^{log_{2}x^4}\]
now we know that \( a^{log_{a} b}=b\)
so we get
\[y= x^4
\]
or
\[x=\sqrt[4]{y}\]

Answer 11

option c

Answer 12

mmm are you sure 2 is the base of log 2x?

Answer 13

in my options i have 4sqrt y, not x=y√4

Answer 14

Let me think

Answer 15

@MelindaR I think 4 sqrt y= \(\sqrt[4] y\)
base has to be 2 otherwise we can't solve this problem.

Answer 16

Oh, the options seem to be always fooling me. Thanks, Ash! you're awesome!

Answer 17

welcome @MelindaR, glad to help:D