# solve the follwing: ln e^3 + ln x^2 = ln (e^5)*x

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ln e^3 + ln x^2 = ln *e^5*x

We know that:

ln a^b = b ln a

==> 3ln e + 2ln x = ln (e^5)*x

Also, we know that ln x+ ln y = ln x*y

==> 3ln e + 2 ln x = ln e^5 + ln x

==> 3ln e + 2 ln x = 5 ln e + ln x

But ln e = 1

==> 3 + 2ln x = 5 + ln x

Now let us group similars:

==> ln x = 2

==> x = e^2

To solve lne^3+lnx^2 = ln(e^5)x

Solution:

By property of logarithms,

lne^3 = 3lne = 3*1 =3

lnx^2 = 2lnx :

ln (e^5)*x = 5lne+lnx = 5*1+lnx = 5+lnx and

Therefore the given equation becomes:

3+2lnx = 5+lnx

2lnx-lnx = 5-3

lnx = 2

x = e^2.

First, we'll use the product property of the logarithmic function, for the right side of the equality:

ln (e^5)*x = ln (e^5) + ln x

Now, we can use the power property of the logarithmic function:

ln e^3 = 3* ln e, but ln e = 1

ln e^3 = 3

ln (e^5) = 5* ln e, but ln e = 1

ln (e^5) = 5

ln x^2 = 2*ln x

We'll re-write the equation:

3 + 2*ln x = 5 + ln x

We'll isolate ln x to the left:

2*ln x - ln x = 5-3

ln x = 2

**x = e^2**