# Properties of Logarithms?

Please help! I've tried and tried. I'm not understanding.

### 6 Answers

- lenpol7Lv 72 months ago
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]

= 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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- Keith ALv 62 months ago
(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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- Wayne DeguManLv 72 months ago
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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- PhilipLv 62 months ago
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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- 2 months ago
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)

Then plug in your values:

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)

Just plug in your values:

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

I hope that helps!

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- BryceLv 72 months ago
a. 2*2 - (4*3 + 2*5)= -18

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

- annabella2 months agoReport
Thank you so much

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