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3 years ago

9

a girl went to school for the first time and asked her how old are you the girl said I don't know but I know my mother's age is

twice my age just asked her how old is your father and the girl said I don't know who my father is is is is is 5 times the age of my mother the girl said I don't know but I know the sum of their ages is 100

1 answer:

elena55 [62]3 years ago

3 0

Answer:

Nw how old is the girl, her mum & her dad?

let x=xmum's age

x+5=dad's age

x/2 =girl's age

x+x+5+x/2=100

3.5x=95

xmum's age=38

dad's age=43

girl's age=19

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2 years ago

URGENT HELP ME PLEASE

Trava [24]

Answer:

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

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

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

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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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3 years ago

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