Question

log x + log 2x = 3 log 2    find x values

Accepted Answer

log x + log 2x = 3 log 2
We know that:
log a + log b = log a*b
==> log x + log 2x = log 2x^2
==> log 2x^2 = 3 log 2
Also, we know that:
a log b = log b^a
==> 3 log 2 = log 2^3 = log 8
==> log 2x^2 = log 8
==> 2x^2 = 8
Now divide by 2:
==> x^2 = 4
==> x= 2

Answer

log x+log2x=3log2
log(x*2x)=log2 to the power 3
log2x^2=log8
x^2=8/2
x=+_2

Answer

We'll start by imposing constraints of existence of logarithms:
x>0
2x>0
The interval of admissible values of x is (0 , +infinite).
Now, we'll solve the equation. We'll start by applying the power property of logarithms:
3 log 2 = log 2^3 = log 8
Now, we'll subtract log 2x both sides:
log x = log 8 - log 2x
We'll apply the quotient rule of logarithms:
log a - log b = log a/b
log x = log 8/2x
log x = log 4/x
Since the bases are matching, we'll apply the one to one property:
x = 4/x
x^2 = 4
x1 = +sqrt4
x1 = +2
x2 = -2
We'll reject the second solution because it doesn't belong to the interval of admissible values.
The only solution of the equation is x = 2.

Answer

logx+log2x= 3log3.
To solve for x.
Solution:
We use the following property of logarithms:
loga+logb =  logab...(1)
And m*loga = loga^m...(2)
So logx+log2x= 3log2.Using property (1), thegiven  equation  becomes logx*3x = 3log2
log3x^2 = log2^3. Here we used the property (2).
Taking antilog, we get:
3x^2 = 2^3 = 8
x^2 = 8/3.
x = sqrt(8/3). Or  x= -sqrt(8/3), but the negative solution is invalid as logarithms are undefined for negatives .

Math

log x + log 2x = 3 log 2 find x values

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