Question

write as a single logarithm...

Answer 1

\[\log_{3}\sqrt{x} -\log_{3}x^{6}\]

Answer 2

the subscript under log is 3

Answer 3

Hello Laurenn ^.^
\[\huge \log_bM - \log_bN = \log_b\frac{M}N\]

Answer 4

i get that, i just don't know where to go from there

Answer 5

Well, why not do it directly? :)
\[\huge = \log_3\frac{\sqrt x}{x^6}\]

Answer 6

But remember \[\huge \sqrt x = x^{\frac12}\]

Answer 7

so would it be \[-1/2\log_{3}x \] ?

Answer 8

Write as a single log first, and then simplify like that... start with this \(\huge =\log_3\frac{\sqrt x}{x^6}\)

Answer 9

how would you simplify that though?

Answer 10

hmm... I guess, you could do this...\[\huge =\log_3\frac{x^{\frac12}}{x^6}\]
and use laws of exponents to express both with just one exponent :)

Answer 11

x^3?

Answer 12

no...
\[\huge =\log_3 x^{\frac12-6}\]

Answer 13

so -5 and a half?

Answer 14

Yup :)
\[\huge = \log_3x^{-\frac{11}2}\]now simplify the way you did :)

Answer 15

yes i got it! -11/2log3x thank you :)

Answer 16

can you help me with other logarithmic equations?

Answer 17

Sure :)

Answer 18

\[2\log_{5}(x-9)+\log_{5}5=3 \]

Answer 19

ohh... solving :)
Here's another property of logs :)
\[\huge \log_bM + \log_b N = \log_bMN\]

Answer 20

But first...
\[\huge 2\log_5(x-9)=\log_b(x-9)^2\]

Answer 21

and then what?

Answer 22

Well, use the property I posted (the log M + log N = log MN)

Answer 23

So...
\[\Large \log_5(x-9)^2 + \log_5 5 = 3\]

Answer 24

Or, actually, you can just remember that
\[\huge \log_bb = 1\]

Answer 25

so log5(x-9)^2+1=3 ?

Answer 26

Yup :)
Now bring everything with no log to the right side ^.^

Answer 27

so 2=log5(x-9)^2 ... would i expand x-9^2 or just leave it like that?

Answer 28

Nah, leave it like that :D
\[\huge \log_bM = x \rightarrow b^x = M\]

Answer 29

So...
\[\huge \log_5(x-9)^2 = 2\]\[\huge (x-9)^2=5^2\]

Don't bother simplifying, just take the square root of both sides ^_^

Answer 30

wait so x-9=square root of 5?

Answer 31

@PeterPan