i will be greatful to you if you help in solving the below problems:
1.)Prove that log15=log5+log3
2.)Show that log10800=4log2+3log3+2log5
3.)show that log108/605=2log2+3log3-log5-2log11.
1.)Prove that log15=log5+log3
hey 15 = 3 * 5
log a b = log a + log b
2.)Show that log10800=4log2+3log3+2log5
log (2^4*3^3*5^2)
log (16*27*25)
log (10800)
3.)show that log108/605=2log2+3log3-log5-2log11.
log [ 2^2*3^3 /(5*11^2) ]
log [ 4*27/(5*121) ]
log [ 108 / 605 ]
all of this is just
log a*b = log a + log b
and
log a^n = n log a
Very thanfulling to u
Sure! I'd be happy to help you solve these problems.
1.) To prove that log15 = log5 + log3, we can start by expressing 15 as a product of 5 and 3.
Using the property of logarithms, we know that log(x * y) = log(x) + log(y).
Therefore, we have log15 = log(5 * 3) = log5 + log3.
2.) To show that log10800 = 4log2 + 3log3 + 2log5, we can first express 10800 in terms of its prime factors.
10800 can be written as 2^4 * 3^3 * 5^2.
Using the property of logarithms, we know that log(x * y) = log(x) + log(y) and log(x^n) = n * log(x).
Therefore, we have log10800 = log(2^4 * 3^3 * 5^2) = 4log2 + 3log3 + 2log5.
3.) To show that log108/605 = 2log2 + 3log3 - log5 - 2log11, we first need to simplify the logarithmic expression.
We can start by expressing 108 and 605 in terms of their prime factors.
108 can be written as 2^2 * 3^3 and 605 can be written as 5 * 11 * 11.
Using the properties of logarithms, we can rewrite the expression.
log108/605 = log(2^2 * 3^3) / log(5 * 11 * 11) = [log(2^2) + log(3^3)] / [log(5) + log(11) + log(11)].
We also know that log(x^n) = n * log(x).
Applying this property, we have [log(2^2) + log(3^3)] / [log(5) + log(11) + log(11)] = [2log2 + 3log3] / [log5 + log11 + log11].
Since we have established that log(x * y) = log(x) + log(y), we can simplify further.
[2log2 + 3log3] / [log5 + log11 + log11] = 2log2 + 3log3 - log5 - 2log11.
Therefore, log108/605 = 2log2 + 3log3 - log5 - 2log11.