Question

i)use the substitution u=2^y to express the equation 3(2^y-1)=2^y+4 as a linear equation of u
ii)hence solve the equation 3(2^y-1)=2^y+4

Answer 1

here u go as u=2^y....
plug in
3(u-1)=u+4
3u-3=u+4
3u-u=4+3
2u=7
u=7/2
now to solve for y
apply natural log
2^y=7/2
log2^y=log(7/2)
ylog2=log7-log2
y=log7-log2/log2=?
or even
y=log7/log4

Answer 2

wait do you mean to solve foy
3(2^y-1)=2^y+4
6^y-3=2^y+4
apply natural log
log6^y-log3=log2^y+log4
log6^y-log2^y=log4+log3
ylog6-ylog2=log4+log3
y(log6-log2)=log4+log3
y=(log4+log3)/(log6-log2).......
not sure how you wanna keep on beaten these you can simplify from there

Answer 3

On the first try, you were ok to here:

y=(log7-log2)/log2
but that's
y = log7log2 - 1

On the 2nd try, you went wrong right away:

3(2^y-1) is NOT 6^y-3
it IS 3*2^y - 3

Answer 4

yeah your are right thanks