Question

Need help with log..1 question..giving out medals

Answer 1

you just want to know what log (1) = ?

Answer 2

please help

Answer 3

\[\Large 2\log_5x+\frac{1}{2}\log_5y-4\log_5z=\log_5\frac{x^2\sqrt{y}}{z^4}\]

So we'll need to use a few rules of logarithms.
\[\large \color{royalblue}{b\cdot\log(a)=\log(a^b)}\]We'll use this rule to deal with the coefficients in front of the logs.

Answer 4

Just for the record, log(1) is not a part of this problem.

Answer 5

\[\Large 2\log_5x=\log_5x^2\]Understand how that works using our rule?

Answer 6

but how do I make it the same?

Answer 7

We're doing that step by step.

Answer 8

I was referring to 1 question not log 1

Answer 9

oh ok..can you show me?

Answer 10

I just showed you the first step above, do you understand so far?

Answer 11

yes I do

Answer 12

\[\Large \frac{1}{2}\log_5y\qquad=\qquad \log_5y^{1/2}\]We can rewrite this using root notation,\[\Large \log_5\sqrt y\]

So using this first rule, what we have so far is,\[\Large \log_5x^2+\log_5\sqrt y-4\log_5z=\log_5\frac{x^2\sqrt{y}}{z^4}\]

Can you follow this idea to simplify this third term?\[\Large 4\log_5z \qquad = \qquad ?\]

Answer 13

log 5z^4?

Answer 14

\[\Large \log_5x^2+\log_5\sqrt y-\log_5z^4=\log_5\frac{x^2\sqrt{y}}{z^4}\]Ok good!
From here we'll need to apply two more rules of logs:\[\large \color{green}{\log(a)+\log(b)=\log(ab)}\]\[\large \color{purple}{\log(b)-\log(c)=\log\left(\frac{b}{c}\right)}\]

When we have the sum of logs, we can rewrite them as a single log with the product of their arguments inside.

Applying this to our first two terms gives us,\[\Large \log_5x^2+\log_5\sqrt y\qquad=\qquad \log_5\left(x^2\sqrt y\right)\]

Answer 15

ok

Answer 16

what is next?

Answer 17

What do you think comes next?

Answer 18

\[\Large \log_5\left(x^2\sqrt y\right)-\log_5z^4=\log_5\frac{x^2\sqrt{y}}{z^4}\]

So next we'll want to combine these two terms,\[\Large \log_5\left(x^2\sqrt y\right)-\log_5z^4 \qquad=\qquad ?\]
using the purple rule I listed.

Answer 19

You should really think about this, because at this point you only have one step and it is very similar to the step he just showed you.

Answer 20

log 5 (x^2 qrt y)/ log 5z^4

Answer 21

thank you!

Answer 22

Close! We don't want to divide logs,\[\Large \log_5\left(x^2\sqrt y\right)-\log_5z^4 \qquad\ne\qquad \frac{\log_5\left(x^2\sqrt y\right)}{\log_5z^4}\]

we want to divide the `insides`.
\[\Large \log_5\left(x^2\sqrt y\right)-\log_5z^4 \qquad=\qquad \log_5\left(\frac{x^2\sqrt y}{z^4}\right)\]

Answer 23

thank you