Question

14log 14 x how would i expand this?

Answer 1

\[\huge 14\log _{14}x\]
is it?

Answer 2

im not to sure i have a pdf it may alivate things

Answer 3

number 9

Answer 4

my teacher makes these hard to read

Answer 5

its \[\huge 14\log _{14}x\]

Answer 6

ok

Answer 7

so hang on

Answer 8

i would first use the power property correct to put the 14 before the log over the x right?

Answer 9

i get that right?

Answer 10

There's nothing to simplify unless making the base 10

Answer 11

So, use the rule
\[\huge \log _{a}b=\frac{\log _{c}b}{\log _{c}a}\]
===========================================
\[\huge 14(\log _{14}x)=14(\frac{logx}{\log14})\]

Answer 12

so change of base?

Answer 13

yep, then
\[14\frac{logx}{\log14}=14\log\frac{x}{14}=14(logx-\log14)\]

Answer 14

okay that makes some sense now on the pdf can you please take a look at number 10?

Answer 15

its simply

\[\log _{14} x^{14} \]

Answer 16

\[\huge \log _9 x=5\]
\[\huge x=9^{5}\]

Answer 17

you jsut changed it into exponential

Answer 18

yes

Answer 19

\[\huge \log _{a}b=c, b=a ^{c}\]

Answer 20

The first formula that you learn logarithms

Answer 21

when*

Answer 22

yeah but it says to find the solution

Answer 23

\[x=9^{5}=59049\]

Answer 24

oh okay that makes sense

Answer 25

@.sam loga/logb does not equal log(a/b)

Answer 26

Guys while you figure out whose right ill be back

Answer 27

@dumbcow i know you will say that XD

Answer 28

so was he right?