How do you solve $5 + log (2 x + 1) = 6$ $5+log(2x+1)=6$ ?

Question

1895 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-16T18:56:21

$x = 4 \frac{1}{2}$ $x = 4 1/2$

When working with equations which contain logs, each term must be a log term. We cannot work with a mixture of logs and numbers.

$5 + log (2 x + 1) = 6$ $5+log(2x+1)=6$

$log (2 x + 1) = 6 - 5$ $log(2x+1)=6 -5$

$log (2 x + 1) = 1$ $log(2x+1)=1$

$log (2 x + 1) = log 10 if log A = log B, then A = B$ $log(2x+1)=log 10" if" log A = log B, " then " A = B$

$2 x + 1 = 10$ $2x+1 = 10$

$2 x = 9 \Rightarrow x = 4 \frac{1}{2}$ $2x = 9 rArr " " x = 4 1/2$

How do you solve 5+log(2x+1)=65+log(2x+1)=65+log(2x+1)=6?