solve these math problems using logs?

Question

2^n=square root of (3^(n-2))

4^x=square root of (5^(x+2))

I know the first step is to square both sides to get rid of the square root and I know logs are used. please show your steps! thanks!

Answer 1

1)
2^n = sqrt(3^(n-2))
2^2n = 3^(n-2) (square both sides)
ln(2^2n) = ln(3^(n-2)) (take natural log of both sides)
2n*ln2 = (n-2)*ln3 (some rule says: ln(x^y) = y*lnx)
2n*ln2 = n*ln3-2*ln3 (expand out right side)
2*ln3 = n*ln3-2n*ln2 (get all n's on one side)
2*ln3 = n*(ln3-2*ln2) (factor out n)
n = 2*ln3/(ln3-2*ln2) (solve for n)
n = -7.638
2)
4^x = sqrt(5^(x+2))
4^2x = 5^(x+2) (square both sides)
2x*ln4 = x*ln5+2ln5 (take log of both sides, use the rule, expand out right side)
x = 2*ln5/(2*ln4-ln5) (factor out x and solve for x)
x = 2.767

Answer 2

Taking log to base 10 of both sides gives:
n log 2 = (n-2) log 3 or
n log 2 - n log 3 = -2 log 3
n(log 2 - log 3) = -2log3
n = -2log(3)/(log(2)-log(3))

The second equation is dealt with similarly:
x log 4 = (x+2) log 5
x log 4 -x log 5 = 2log 5
x = 2 log 5 /(log 4 - log 5)

Answer 3

2^n= V[3^(n-2)] = [3^(n-2)]^(1/2)
Take logs of both sides, as long as n is real this is applicable since for real x, log x is one-one. We obtain,
log 2^n = log [3^(n-2)]^(1/2)
n.log 2 = (1/2).log [3^(n-2)] = (1/2).(n-2).log 3
then,
2n.log 2 = n.log 3 - 2.log 3
which gives,
n(log 3 - 2.log 2) = 2.log 3
n.[log (3/2^2)] = 2.log 3
n = 2.log 3 / log (3/4)

Note that m.log b = log (b^m)

Answer 4

log then : n = n/2-1 then n+1 = n/2 then 2*n +2 =n then n= -2
log then x = x/2+1 then x-1 = x/2 then 2*x-2 = x then x= 2

Answer 5

we are going to initiate with the help of taking the organic logarithm of both area. (i will use "ln" to face for the organic log.) ln [ 2e^(3x) ] = ln [4e^(5x) ] utilizing the houses of logarithms, ln 2 + ln [ e^(3x) ] = ln 4 + ln [ e^(5x) ] ln 2 + 3x ln e = ln 4 + 5x ln e ln 2 + 3x = ln 4 + 5x (because ln e = a million) Simplifying: ln 2 - ln 4 = 5x - 3x ln 2 - ln (2²) = 2x ln 2 - 2 ln 2 = 2x (utilizing the houses of logs back, on the left-hand area) - ln 2 = 2x . . . . ln 2 x = - ------- . . . . . 2 Sorry about the strange formatting, yet i necessary to make the fraction line up and Yahoo solutions is somewhat clumsy about that. besides, desire that facilitates!

Answer 6

2^n=square root of (3^(n-2))
n ln 2=1/2 ln 3^(n-2)
n ln 2=(n-2)/2 ln 3
n ln 2=n/2 ln 3-ln 3
n(ln 2-.5 ln 3)=-ln 3
n=-ln 3 (ln 2-.5ln 3)=-1.0986/(0.6931-.5493)=-7.638

4^x=square root of (5^(x+2))
x ln 4=(x+2)/2 ln 5
x ln 4=x/2 ln 5 +ln 5
x(ln 4 -.5 ln 5)=ln 5
x=ln 5 /(ln 4-.5 ln 5)
x=1.6094/(1.386-.805)=2.767