Question

2+log2(2x−2)=2log2(x−1)
the (log2),the 2 should be a bit down the log

Answer 1

2 is the base of log ?

Answer 2

\[2+\log_{2} (2x-2)=2\log_{2}(x-1) \]

Answer 3

right?

Answer 4

4(2x-2)=(x-1)(x-1)

Answer 5

start with
\[ 2=\log_2((x-1)^2)-\log_2(2(x-1))\]then
\[2=\log_2(\frac{(x-1)^2}{2(x-1)})\] then
\[2=\log_2(\frac{x-1}{2})\] and finally
\[\frac{x-1}{2}=2^2=4\] and solve for x

Answer 6

actually jhonny method is snappier!

Answer 7

but still get
\[x-1=8\]
\[x=9\]

Answer 8

thanks i got 9 as my answer too