help me plz
- Thread starter janekela
- Start date
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
First step is to take the log of both side.
Then use the property of logs such that:
log(a*b) = log(a) + log(b)
and
log(x^y) = ylog(x)
you should end up with x = 1/2
Then use the property of logs such that:
log(a*b) = log(a) + log(b)
and
log(x^y) = ylog(x)
you should end up with x = 1/2
Hello, janekela!
We don't need logs . . .
We have: \(\displaystyle \
4^x)\cdot(5^{2x})\;=\;10\)
The left side is: \(\displaystyle \:[(2^2)^x]\cdot(5^{2x})\;=\;(2^{2x})(5^{2x})\;=\;(2\cdot5)^{2x}\;=\;10^{2x}\)
The equation becomes: \(\displaystyle \:10^{2x}\;=\;10^1\)
Therefore: \(\displaystyle \;2x\,=\,1\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\)
We don't need logs . . .
We have: \(\displaystyle \
4^x)\cdot(5^{2x})\;=\;10\)The left side is: \(\displaystyle \:[(2^2)^x]\cdot(5^{2x})\;=\;(2^{2x})(5^{2x})\;=\;(2\cdot5)^{2x}\;=\;10^{2x}\)
The equation becomes: \(\displaystyle \:10^{2x}\;=\;10^1\)
Therefore: \(\displaystyle \;2x\,=\,1\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\)