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3 years ago

5

25 POINTS! Which equation is equivalent to (1/3)=27^x^+^2

2 answers:

GenaCL600 [577]3 years ago

8 0

Answer:

match bases

remeber if a^b=a^c where a=a then b=c

remember

(x^m)^n=x^(mn)

1/(x^m)=x^-m

base of 3

(1/3)^x=(3^-1)^x=3^(-x)

27^(x+2)=(3^3)^(x+2)=3^(3x+6)

3^(-x)=3^(3x+6)

-x=3x+6

0=4x+6

-6=4x

-3/2=x

Genrish500 [490]3 years ago

7 0

Hint: $\bf log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}$

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3 years ago

Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

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Considering the given equation $y = log_{3}x\\$

And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For $(\frac{1}{3}, a_{0} )$

$x=\frac{1}{3}$

$y=a_{0}$

$y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1$

So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

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3 years ago