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Troyanec [42]
3 years ago
9

Please Help Asap thank you!!​

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
6 0
Volume 68892837 volume 64929 could 7383 could 9
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If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
3 years ago
12,13,14 help please
PSYCHO15rus [73]
12. 15 (3x5x1)
13.  8  (4x2x1)
14. 18 (6x3x1)

Hope this helps! :D
5 0
3 years ago
Read 2 more answers
Division of whole numbers 3075 divided by 3
butalik [34]

Answer:

1025

Step-by-step explanation:

3075/3=1025

4 0
3 years ago
The distance between the points (-1, -6) and (-19, -6) is units
Vladimir [108]

Answer:

18

Step-by-step explanation:

\sqrt{  {( - 9 -  - 1)}^{2} +  {( - 6 -  - 6)}^{2} }

\sqrt{ {( - 18)}^{2} }

\sqrt{324}

3 0
3 years ago
Explain number 2 &amp; 3 please
VLD [36.1K]
1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25
7 0
3 years ago
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