# Properties of Logarithms? Relevance
• ln(a^2/b^4c^2)

All the 'stuff' inside the brackets are multiplications/quotients.

When you split this up in logarithmic form , they become addition/subtraction.

Hence the above statement becomes

ln a^2 - ln b^4 - ln c^2

Another condition of logs. is that if the component under the 'log' function is raised to a power, then the power can be removed to become a coefficient of the log function

hence

ln a^2 = 2ln a

Similarly

ln b^4 = 4 ln b

ln c^2 = 2 ln c

Hence

ln(a^2/b^4c^2) = 2 ln a - 4 ln b - 2 ln c

Substitute in the number from the question

2(2) - 4(3) - 2(5) = 4 - 12 - 10 = - 18

(b)

ln(a^1b^3c^4)^(1/2)

NB The surd(root) sign can be changed to the power of '1/2'  (square root).

Hence

(1/2) ln(a^1b^3c^4) =>

(1/2)[ (1) ln a + 3 ln b + 4 ln c]

Substituting again

(1/2) [ 1(2) + 3(3) + 4(4)]

(1/2)[ 2 + 9 + 16]  =>

(1/2)

= 27/2  = 13 1/2 = 13.5

NB

Remember when calculating with exponents

a^2 X a^5 = a^(2 + 5) = a^7  Notice that when multiplying two terms with the same coefficient, then you add the exponents. Logarithms are the inverse of this process.

Then if a^7 = b

Then log(a)b = 7

NNB  you have used base 'e' (ln) in the question.

You change the log base, but that is a whole new calculation.

NNNB Your calculator will be programmed for logs base 'e' (ln) and separately for logs base '10' (log). Do not casually interchange the base, but stick with the same base.

Hope that helps!!!!!

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• (If you are using real numbers, then the argument of the logarithm must be positive.)

Think this, first:

"A logarithm is an index (is an exponent)"

The three words are merely used in different contexts, and with different emphases.

If you want to prove, or verify, some property of logarithms, first give a name to one or two of the logarithms, then translate to index form:

E.g.:

log (x*y) = log x + log y  --  true for any base.

The base needs to positive and not  1.  Call it  a.

Let  u = log[a] (x),  and  v = log[a] (y)

That means  (with  a the base, and  u  or  v  the index)  x = a^u , and  y = a^v./

Now  x*y = a^u * a^v = a^(u + v)

Translate back:

log (x*y) = u + v = log x + log y.

Similarly for log (x / y);

and  log (x^n);

(and some less easy  ones).

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• ln(A/B) => lnA - lnB

so, ln(a²/b⁴c³) = ln(a²) - ln(b⁴c³)

Also, ln(AB) => lnA + lnB

so, ln(a²) - [ln(b⁴) + ln(c³)]

i.e. ln(a²) - ln(b⁴) - ln(c³)

And ln(Aⁿ) => nlnA

so, 2ln(a) - 4ln(b) - 3ln(c)

Hence, 2(2) - 4(3) - 3(5) => -23

ln√(ab³c⁴) => (1/2)ln(ab³c⁴)

i.e. (1/2)(lna + lnb³ + lnc⁴)

so, (1/2)(lna + 3lnb + 4lnc)

Then, (1/2)(2 + 3(3) + 4(5))

=> (1/2)(31) = 15.5

:)>

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• ln(a,b,c) = (2,3,5).

(a) ln[(a^2)/(b^4c^2)] = 2ln[a/(cb^2)] =;2{lna-lnc-2lnb} = 2(2-5-2*3) = 2(-9) = -18.

(b) ln[(a^1)(b^3)(c^4)]^(1/2) =;

(1/2)[lna +3lnb +4lnc]= (1/2)[2 +3*3 +4*5];

= (1/2)(2+9+20) = (31/2) = 15.5.

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• There's a few properties of logarithms you'll need to use here:

a*ln(x) = ln(x^a)

ln(x/y) = ln(x) - ln(y)

ln(xy) = ln(x) + ln(y)

Looking at the first one, we can first apply the second property I have listed and then the third:

ln(a^2/b^2c^2) = ln(a^2) - ln(b^2c^2) = ln(a^2) - ln(b^2) - ln(c^2)

Then we can use the first one:

2ln(a) - 2ln(b) - 2ln(c)

2 - 3 - 5 = -6

For the second, it will help if we first distribute our square root into each exponent:

ln(a^(1/2)*b^(3/2)*c^2)

Then use the third property to break them up:

ln(a^(1/2)) + ln(b^(3/2)) + ln(c^2)

And finally the first property:

(1/2)ln(a) + (3/2)ln(b) + 2ln(c)

(1/2)(2) + (3/2)(3) + 2(5) = 1 + 9/2 + 10 = 31/2

I hope that helps!

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• a. 2*2 - (4*3 + 2*5)= -18

b. (1/2)(2 + 3*3 + 4*5)= 31/2= 15.5

• Thank you so much

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