: Solve:
(2^x)-(3^x)=0
I tried to bring it to base 2 but it didn't work. Please show us steps.
Thank you.
2x - 3x = 0
2x = 3x
There are two log bases on your
calculator, "log" which is the
logarithm base 10, and "ln" which
is the natural logarithm base "e".
You can use either one:
If we use "log", we take the log | If we use "ln", we take the ln
of both sides: | of both sides:
log(2x) = log(3x) ln(2x) = ln(3x)
|
Now we use the rule of logs to | Now we use the rule of lns to
rewrite both sides: | rewrite both sides:
|
log(AB) = B·log(A) ln(AB) = B·ln(A)
|
x·log(2) = x·log(3) | x·ln(2) = x·ln(3)
|
Get 0 on the right | Get 0 on the right
|
x·log(2) - x·log(3) = 0 | x·ln(2) - x·ln(3) = 0
|
Factor out x on the left: | Factor out x on the left:
|
x·[log(2) - log(3)] = 0 | x·[ln(2) - ln(3)] = 0
|
Using the calculator: | Using the calculator:
|
x·{.301 - .477) = 0 | x·(.693 - 1.099) = 0
|
x·(-.176) = 0 | x·(-.406) = 0
|
-.176x = 0 | -.406x = 0
|
Divide both sides by -.176 | Divide both sides by -.406
|
x = 0 | x = 0
|
Either way you get x = 0
Checking:
2x - 3x = 0
20 - 30 = 0
1 - 1 = 0
0 = 0
Edwin