Question

Solve the equation of

Answer 1

\[\frac{ 1 }{ 4} \log_{6} (a-3)-\log_{6} 3=0\]

Answer 2

First step, we can get rid of the fraction by multiplying each term by 4.

Then we can make us of the following identities: \[\Large n * \log(A) = \log(A ^{n})\] and \[\Large \log(A) - \log(B) = \log(\frac{ A }{ B })\]and if log(Q) = 0 then Q = 1.

Answer 3

ok let me see

Answer 4

would the first one be log6^1/4

Answer 5

The (a-3) will be raised to 1/4 not the 6.

Answer 6

and log3^-1

Answer 7

i mean log -1/3

Answer 8

That step is not necessary. All it does is turn the minus sign to a plus sign in front of the second log. But it is not necessary. You can use the formula that the difference of two logs: log(A) - log(B) = log(A/B)

Answer 9

to solve this would it be eayer to use log(A/B)

Answer 10

There is an even easier way to solve this particular problem, but if you use the log(A/B) method it may come in handier for other problems.

Answer 11

should i do the a-3^1/4 first ?

Answer 12

Yes.

Answer 13

i can put 1/(a^4-81)/1/3

Answer 14

Not sure how you arrived at that result. It is incorrect.

Answer 15

hmm let me try again

Answer 16

for (a-3)^1/4 isnt that 1/(a^4-81

Answer 17

It is not to the 4th power. It is 1/4. Also no need to expand (a-3)^1/4. It can be left as it is.

Answer 18

\[(a-3)^{\frac{ 1 }{ 4}}-\frac{ -1 }{3 }\]

Answer 19

is that the equation?

Answer 20

I havent done anything like this before XD

Answer 21

I will show the simplest way first:
\[\Large \frac{ 1 }{ 4 }\log_{6} (a-3) - \log_{6} 3 = 0\]\[\Large \log_{6} (a-3)^{\frac{ 1 }{ 4 }} - \log_{6} 3 = 0\]. This is possible only when \[\Large (a-3)^{\frac{ 1 }{ 4 }} = 3\]. Raise both sides to the 4th power and solve for a.

Answer 22

a=84?

Answer 23

Yes a = 84.

Answer 24

thanks again i get i tnow

Answer 25

cool. so we won't need the longer method, right?

Answer 26

right

Answer 27

Thats way way easyer

Answer 28

alright.