SOLUTION: Please help. Find the real solution to: {{{log(2)t+log(2)(t+1)=1}}} Answer: t=1 But how? I have tried: {{{2^(1)+2^(t+1)=t^2+t}}} {{{2^(x+1)=e^(2)+t}}} and {{{log(2)t+l

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help. Find the real solution to: {{{log(2)t+log(2)(t+1)=1}}} Answer: t=1 But how? I have tried: {{{2^(1)+2^(t+1)=t^2+t}}} {{{2^(x+1)=e^(2)+t}}} and {{{log(2)t+l      Log On


   

Question 981943: Please help. Find the real solution to:
log%282%29t%2Blog%282%29%28t%2B1%29=1
Answer: t=1
But how? I have tried:
2%5E%281%29%2B2%5E%28t%2B1%29=t%5E2%2Bt
2%5E%28x%2B1%29=e%5E%282%29%2Bt
and
log%282%29t%2Blog%282%29t%2Blog%282%291=1
Answer by josgarithmetic(39626) About Me  (Show Source):
You can put this solution on YOUR website!
Your equation shown is ambiguous. One would only try guessing the base is meant to be 2.

log%28%28t%29%29%2Blog%28%28t%2B1%29%29=1, with the base of 2.

log%28%28t%28t%2B1%29%29%29=1

2%5E1=t%28t%2B1%29

t%5E2%2Bt=2

t%5E2%2Bt-2=0

%28t-1%29%28t%2B2%29=0

2 to no power gives any negative value, so you can only accept t=1.