Solve for x:
log2x+1=log4x
I know the answer is 1/4 but I don't know how to get there. Help please.
+129839 If we have log(2x+1)=log(4x). we can just set 2x + 1 = 4x and solve
So we have
2x + 1 = 4x
1 = 2x
x = 1/2
------------------------------------------------------------------------------------------------
If we have log(2x)+1=log(4x) .....we have (writing "1" as "log 10" )
log(2x)+ log 10 = log4x and by a property of logs, we can write
log(2x*10) = log(4x)
log (20x) = log(4x)
And setting 20x = 4x → x = 0 ......but this answer isn't possible because we would be taking the log of 0 and that's undefined......so..... log(2x)+1=log(4x) has no solution....
The graph of the second situation can be seen here.....https://www.desmos.com/calculator/cdqfqzhvpa
Note that the graphs never intersect......thus, there are no solution points......
+129839 If we have log(2x+1)=log(4x). we can just set 2x + 1 = 4x and solve
So we have
2x + 1 = 4x
1 = 2x
x = 1/2
------------------------------------------------------------------------------------------------
If we have log(2x)+1=log(4x) .....we have (writing "1" as "log 10" )
log(2x)+ log 10 = log4x and by a property of logs, we can write
log(2x*10) = log(4x)
log (20x) = log(4x)
And setting 20x = 4x → x = 0 ......but this answer isn't possible because we would be taking the log of 0 and that's undefined......so..... log(2x)+1=log(4x) has no solution....
The graph of the second situation can be seen here.....https://www.desmos.com/calculator/cdqfqzhvpa
Note that the graphs never intersect......thus, there are no solution points......
