Math Problem, Natural Logs....?

Question

Even though they may be called natural logs, these logs don't come natural to me. **Hits Drum Symbol**. Please help me.

Solve for X:
ln x + 2ln4 = ln7 - ln x

Answer 1

ln(x) + 2ln(4) = ln(7) - ln(x)
ln(x) + ln(4^2) = ln(7) - ln(x)
ln(x*4^2) = ln(7/x)
ln(16x) = ln(7/x)
16x = 7/x
16x^2 = 7
x^2 = 7/16
x = sqrt(7/16)
x = sqrt(7)/4

Answer 2

The sum of logs is the product of a single log; difference is quotient; a coefficient becomes an exponent, so rewrite as singlr ln's:

ln((x)(4^2) = ln (7/x)

ln(16x) = ln (7/x)

so 16x must = 7/x; 16x^2 = 7; x^2 = 7/16; x = +- sqrt7 / 4

Can't take the log of a negative number, so just sqrt7 / 4 is the answer

Answer 3

ln x+2ln4= ln7- ln x
2lnx=ln7-ln4^2
lnx^2=ln7/16
x^2=7/16
x=positive sq. root of7/16

Answer 4

ln x + 2ln4 = ln7 - ln x

lnx + ln16 = ln(7/x)

ln(x*16) = ln(7/x)


16x = 7/x

x^2 = 7/16

x = √(7)/4

Learn your laws of logs and you'll find these are easier!

Answer 5

ln x + 2ln4 = ln7 - ln x
So 2lnx = ln7 - 2ln4
ie ln(x^2) = ln(7/(4^2)) (as alnb = ln(b^a) and lnp - lnq = ln(p/q))
ie x^2 = 7/(4^2)
So x = √7 / 4

Answer 6

u don't use calculators??

transpose u got 2lnx= 2ln4-ln7
lnx= (2ln4-ln7)/2
inverse both ln
x= ln^- [(2ln4-ln7)/2]