Question

Write the expression as a single logarithm.

Answer 1

\[5\log_{b} y+6\log_{b} x\]

Answer 2

There are three basic but very important rules of logs that you should know well:

ln x + ln y = ln (x*y)
ln x - ln y = ln (x/y)
ln x^y = y*ln x

which of these do you think applies to your math problem? Go ahead and apply them.

Answer 3

The first one?

Answer 4

Actually, Cassidy, you'll need more than 1 rule, but fewer than 3 rules! :) Try applying them; show me your work if at all possible. Maybe use the Draw feature, below?

Answer 5

5\log_{b} y+6\log_{b} x=\log_{b} y ^{5}+\log_{b} x ^{6}

Answer 6

The equation isn't working :(

Answer 7

I'm not sure why it did that

Answer 8

See below (typing):

Answer 9

\( \bf {5log_b(y)+6log_b(x)\qquad {\color{red}{ a}}log_b(c)\implies log_b(c^{\color{red}{ a}})\\ \quad \\

log_b(y^{\color{red}{ \square }})+6log_b(x^{\color{red}{ \square }})\qquad log(a)+log(b)\implies log(a\cdot b)\\ \quad \\

log_b(\square? \cdot \square? )}\)

Answer 10

\[5\log_{b} y = \log_{b} y ^{5}\] Why? Which rule did we apply here?

Answer 11

y^5x^6?

Answer 12

The third rule?

Answer 13

Work on the two separate terms of the given expression separately.

In this example
\[5\log_{b} y=\log_{b} y ^{5}, \]which rule did we use?
Answer: the third rule.

Answer 14

Process the 2nd term in the same way. Then you'll have two logs added together. Use the first rule to simplify that. Try again, and please show us your actual work if at all possible. Can give you better feedback that way.

Answer 15

hmmm I even got a typo... \(\bf {5log_b(y)+6log_b(x)\qquad {\color{red}{ a}}log_b(c)\implies log_b(c^{\color{red}{ a}})\\ \quad \\

log_b(y^{\color{red}{ \square }})+log_b(x^{\color{red}{ \square }})\qquad log(a)+log(b)\implies log(a\cdot b)\\ \quad \\

log_b(\square? \cdot \square? )}\)

Answer 16

any ideas?

Answer 17

Thanks, Joe Doe.
Mind guiding Cassidy through the transformation of the 2nd term,
\[6\log_{b} x\]??

Answer 18

sure

Answer 19

\(\bf {\color{red}{ 5}}log_b(y){\color{red}{ 6}}log_b(x)\qquad\qquad {\color{red}{ a}}log_b(c)\implies log_b(c^{\color{red}{ a}})\\ \quad \\

log_b(y^{\color{red}{ 5 }})+log_b(x^{\color{red}{ 6 }})\qquad\qquad log(a)+log(b)\implies log(a\cdot b)\)

Answer 20

then just use the rule of \(\bf log(a)+log(b)\) and you'd get \(\large \bf log_b(\square \cdot \square )\)

Answer 21

woops... that is the rule of \(\bf log(a)+log(b)\implies log(a\cdot b)\)