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Anuta_ua [19.1K]

2 years ago

5

I'm stuck on this one. I'm sorry of you know everything but there are other people who can't comprehend some problems without be

ing their to actually see and experience it with a teacher. Just help me with this one -

1 answer:

Gekata [30.6K]2 years ago

7 0

Answer:

Expand the brackets, and simplify.

(4t - 8/5)-(3-4/3t) = (4t +4/3t) + ( -8/5 - 3) = 5 1/3t - 23/5 = 16/3t - 23/5

5(2t + 1) + (-7t + 28) = 10t + 5 - 7t + 28 = 3t + 33

(-9/2t + 3) + (7/4t + 33) = (-9/2t + 7/4t) + (3 + 33) = -11/4t + 36

3(3t - 4) - (2t + 10) = 9t - 12 - 2t - 10 = 7t - 22

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URGENT HELP ME PLEASE

Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(c) $\log_2(\dfrac{64}{8})=3$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

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The answers
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The answer is A!

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