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
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
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
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
yeah your are right thanks
To express the equation 3(2^y-1) = 2^y + 4 as a linear equation of u, we'll follow these steps:
i) Start by substituting u = 2^y.
Substituting u into the equation, we get: 3(u - 1) = u + 4.
ii) Simplify the equation.
Multiplying 3 by u and 3 by -1, the equation becomes: 3u - 3 = u + 4.
iii) Rearrange the equation to isolate u.
To isolate u, let's move all terms with u to one side of the equation: 3u - u = 4 + 3.
iv) Combine like terms.
Simplifying, we have: 2u = 7.
v) Solve for u.
Dividing both sides of the equation by 2 gives us: u = 7/2.
Therefore, the linear equation that expresses the equation 3(2^y-1) = 2^y + 4 in terms of u is 3u - 3 = u + 4, and the solution to this equation is u = 7/2.