-5=log2X
-5 = log2x
2^-5 = x
x = 1/32
3^(5x+1)= 40
let's take the log of both sides
log(3^(5x+1)) = log40
(5x + 1) log3 = log40
5x+1 = log40/log3
etc.
log6X+(log6)7=(log6)35
log6(7x) = log6 35
"antilog" both sides
7x = 35
etc.
Can someone please explain to me how to solve logarithmic equations? Some examples are: -5=log2X
3^5x+1= 40 and log6X+(log6)7=(log6)35
I hope that makes sense because logs are hard to type.
1 answer
1 answer