Question

log2x-log2(x-2)=3 find x

Answer 1

Use this rule of logs first.

\( \large \log_b x - \log_b y = \log_b \dfrac{x}{y} \)

Answer 2

@666slayer are you stuck here?

Answer 3

yes the answer to x is 16/7 but i don't know how to work the problem to get the answer.

Answer 4

Okay, doing as my colleague suggested:
\[\log_2 x - \log_2 (x-2) = 3\]\[\log_2 (\frac{x}{x-2}) = 3\]

Answer 5

Do you see how to do it now?

Answer 6

Raise the base of the logarithm to both sides…\[2^{\log_2 (\frac{x}{x-2})} = 2^3\]\[b^{\log_b u} = u\]

Answer 7

2log2(x/x-2)=8 where do you go from here?

Answer 8

\[2^{\log_2 (\frac{x}{x-2})} = 2^3\]But we apply that principle \[b^{\log_b u} = u\]and we get
\[\frac{x}{x-2} = 8\]You can solve that, right?

Answer 9

\[x = 8(x-2)\]\[x=8x-16\]\[-7x=-16\]\[x=\frac{16}{7}\]

Answer 10

Thanks a lot i got it now