# 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

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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

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**

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

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.

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.