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in `2 log x- log 4= 2` solve for x

in `2 log x- log 4= 2` solve for x

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in `2 log x- log 4= 2` solve for x

Postby Galaxy » Tue 12 April, 2011 18:14

how can I solve for x in in `2 log x- log 4= 2` ?
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Re: in `2 log x- log 4= 2` solve for x

Postby barnamah » Tue 12 April, 2011 18:24

to solve for x in `2 log x- log 4= 2` we can simplify it so we can find X.
for the term `2logx` we can wire it as `logx^2` putting it together we have:
`logx^2-log4=2`

Now if we have `logA-logB` we can write it as `log(A/B)` (we have log with the same base 10 here) applying this rule we have:
`log(x^2/4)=2`
to solve for x we should know that `logA=B` to find A is `10^B=A` . Remember when log does not have the base such as `log_(5)x=3` means it is based on 5. In our case it is based 10 and we don't write base 10 because everyone knows that.


Applying this rule we can write `10^2=x^2/4` and now we got rid of log, we can solve for x.
Simplify for `10^2=100` we can write `100=x^2/4`
Multiply both side by 4 we have: `4*100=x^2/4*4` the 4 on right cancel out and we got:
`400=x^2`. Now solve for x, we take the square of root of both side:
`sqrt400=sqrt(x^2)` the square root and root cancel out and we have left with x on the right.
`sqrt400=x`

so `x=sqrt400=20`
x=20
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