Question

Simplify: i) ln(2x) + ln(2/x) ii) ln(x^2 - 1)- ln(x-1) Then solve: i) lnx = 5 ii) lnx + ln5 = ln10Simplify:
i) ln(2x) + ln(2/x)
ii) ln(x^2 - 1)- ln(x-1)
Then solve:
i) lnx = 5
ii) lnx + ln5 = ln10

Accepted Answer

i) ln(2x) + ln(2/x)
We know that:
ln a + ln b = ln a*b
==> ln (2x)*(2/x) = ln 4
 
ii) ln(x^2 - 1)- ln(x-1)
We know that:
ln a - ln b = ln (1/b)
==> ln (x^2-1)/(x-1) = ln (x+1)
 
i) lnx = 5
==> x= e^5
 
ii) lnx + ln5 = ln10
==> ln 5*x = ln 10
==> 5x = 10
==> x= 2

Answer

You need to simplify  using the logarithmic identities such that:



You need to simplify  using the logarithmic identities such that:



You need to solve for x the equation  such that:

Hence, evaluating the solution to the given equation yields  
You need to solve for x the equation  such that:


Hence, evaluating the solution to the given equation yields

Answer

Here we have to simplify:
ln(2x) + ln(2/x) and ln(x^2 - 1)- ln(x-1)
We use the relation that ln A + ln B = ln A*B
and ln A - ln B = ln (A / B)
ln(2x) + ln(2/x)
=> ln(2x) + ln(2/x)
=> ln ( 2x * 2/x)
=> ln 4
ln(x^2 - 1)- ln(x-1)
=> ln ( x^2 - 1) / (x - 1)
=> ln (x+1)*(x-1)/(x-1)
=> ln (x+1)
Now to solve the two equations for x:
ln x = 5 ,
=> x = e^5
ln x = ln 10 - ln 5 
=> ln x = ln ( 10/5 )
=> ln x = ln 2
=>  x = 2

Answer

To simplify
i) ln(2x) + ln(2/x)
ii) ln(x^2 - 1)- ln(x-1)
(i)
ln(2x)+ln(2/x) = ln(2x)(2/x), as lna+lnb = ln(ab).
ln(2x)+ln(2/x) =ln(4x/x) = ln4.
l(ii) ln(x^2-1)-ln(x-1) = ln (x^2-1)/x-1) , as lna-lnb = ln(a/b)
ln(x^2-1) - ln(x-1) = ln(x-1)(x+1)/(x-1) , as a^2 -b^2 = (a-b)(a+b).
ln(x^2-1) - ln(x-1) = ln(x+1).
Then solve:
i) lnx = 5
lnx = 5.
x = e^5, as lif ln x = b , the x = e^b by definition of logarithm.
ii) lnx + ln5 = ln10
lnx +ln5 = ln10
ln x = ln10 - ln5
lnx = ln(10/5), as lna-lnb = ln (a/b).
lnx = ln2.
Take antilog:
x = 2.

Answer

To simplify the given expressions, we'll apply the product and quotient rules of logarithms.
1) ln(2x) + ln(2/x)
We'll apply the product rule,because the bases of the logarithms are matching, so, we'll transform the sum of logarithms into the logarithm of the product.
ln(2x) + ln(2/x) = ln (2x*2/x)
We'll reduce like terms and we'll get:
ln(2x) + ln(2/x) = ln 4
2) ln(x^2 - 1)- ln(x-1)
We'll apply the quotient rule: The differenceof logarithms is the logarithm of the quotient.
ii) ln (x^2 - 1) - ln (x-1) = ln [ (x^2 - 1)/(x-1)]
We notice that we have a difference of squares, at numerator:
 (x^2 - 1) = (x-1)(x+1)
ln [ (x^2 - 1)/(x-1)] = ln [ (x-1)(x+1)/(x-1)]
We'll simplify the ratio:
ln [ (x-1)(x+1)/(x-1)] = ln (x+1)
Now, we'll solve the next 2 equations:
1) lnx = 5
We'll write the right side as 5*1 = 5*ln e
lnx = 5*ln e
We'll use the power rule of logarithms:
5*ln e = ln e^5
ln x = ln e^5
Because the bases are matching, we'll use the one to one property:
x = e^5
2) lnx + ln5 = ln10
We'll apply the product rule,because the bases of the logarithms are matching.
lnx + ln5 = ln 5x
ln 5x = ln 10
Because the bases are matching, we'll use the one to one property:
5x = 10
We'll divide by 5:
x = 2
Because the value of x is positive, the solution is valid.

Math

Simplify: i) ln(2x) + ln(2/x) ii) ln(x^2 - 1)- ln(x-1) Then solve: i) lnx = 5 ii) lnx + ln5 = ln10 Simplify: i) ln(2x) + ln(2/x) ii) ln(x^2 - 1)- ln(x-1) Then solve: i) lnx = 5 ii) lnx + ln5 = ln10

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