Question

Impossible?

Part 1 Is the following equation true?

Part 2 Use complete sentences to explain the properties used in making your decision.

log 3 x + 1/2 log 3 y – 2 log 3 z = log 3 x^4√y/z^2

Answer 1

This is what I found but I don't understand it

This equation is true.

4log[3] x + (1/2)log[3] y - 2log[3] z
= log[3] x^4 + log[3] y^(1/2) - log[3] z^2
= log[3] (x^4)(sqrt(y)) - log[3] z^2
= log[3] [(x^4)(sqrt(y)) / z^2]

Properties used:
a log b= log b^a
log a + log b = log (ab)
log a - log b = log (a/b)
a^(1/2) = sqrt(a)

Answer 2

\[\log_{3}x+\frac{1}{2}\log_{3}y-2\log_{3}z=\log_{3}x^4\frac{y}{z^2} \]

Answer 3

Is this the problem?

Answer 4

x^4 should be on top with y

Answer 5

\[\frac{1}{2}\log_{3}y=\log_{3}\sqrt{y} \]

Answer 6

\[2\log_{3}z=\log_{3}z^2 \]

Answer 7

So now we have:
\[\log_{3}x=\log_{3}\sqrt{y}-\log_{3}z^2=\log_{3}\frac{x^4y}{z^2} \]

Answer 8

Whoops...make that a plus sign

Answer 9

log a + log b = log ab so we have:
\[\log_{3}x \sqrt{y}-\log_{3}z^2=\log_{3}\frac{x^4y}{z^2} \]

Answer 10

Now if we add
\[\log_{3}z^2 \]
to both sides we have:
\[\log_{3}x \sqrt{y}=\log_{3}\frac{x^4y}{z^2}+\log_{3}z^2 \]

Answer 11

But using log a + log b = log ab on the right side gives us:
\[\log_{3}x \sqrt{y}=\log_{3}x^4y \]

Answer 12

\[0=\log_{3}x^4y-\log_{3}x \sqrt{y} \]

Answer 13

Doesn't look true to me.

Answer 14

in the original problem did you make the x^4 so it was with square root/ z^2? @Mertsj

Answer 15

*square root of y