Question

Solve for x:
log3 sqrt x+1 - log1/3 1/(x+1)^2 = log9 64

Answer 1

Could you please put more parentheses to show how the question looks exactly?

Answer 2

\[\log_{3}\sqrt{x+1} -\log_{1/3}\frac{ 1 }{ (x+1)^{2} }=\log_{9}64\]

Answer 3

@allank. Thanks! :))

Answer 4

No prob. I would first convert all the logarithms to use the same base...know how to do that?

Answer 5

that's what i need to know :((

Answer 6

you can not evaluate unless the bases are the same :(((

Answer 7

It's quite simple. Here's an example of base 3 to base 10:
\[\log_{3} 5 = \log_{10}5/\log_{10}3 \]

Answer 8

Another example:
\[\log_{7}32=\log_{10}32/\log_{10}7 \]

Answer 9

See the trend?

Answer 10

Yes, always with the base 10. Then the original base itself becomes the number for the log

Answer 11

am i right?

Answer 12

Almost. This works for all bases, not just base 10. One more example of base 9 to base 2:
\[\log_{9}87 = \log_{2}87/\log_{2}9 \]

Answer 13

how would i know what the new base to use?

Answer 14

That's your choice...unless specified. In this question, whichever base you choose will give you the correct answer. So you probably want to choose a base that will make life easy for you...namely base 10, because a calculator will be able to compute the logarithms for you.

Answer 15

any value for the base. COOL :)) what about if the base is a fraction?

Answer 16

Same method applies:
\[\log_{1/3}25 = \log_{10}25/\log_{10}(1/3) \]

Answer 17

Thanks. How do you go about now with the solving for x?

Answer 18

will i use the Properties of log?

Answer 19

Yep, after doing some simplification. I choose base 10 then convert everything to that base to get:
\[\log_{10}(\sqrt{x+1})/0.4771 - \log_{10}(1/(x+1)^2)/-0.4771 = \log_{10}64/\log_{10}9 \]

Answer 20

Can you spot how I used base change?

Answer 21

I believe so. Thus, you can combine the two terms on the left of the equation, then use log properties to combine the logs in the numerator and so on an so forth. I hope that helps. :)

Answer 22

THANK YOU VERY MUCH @allank . YOU'RE SUCH A BIG HELP :)))

Answer 23

You're very welcome :)