Question

Help to solve one equation?
23^x = 6
I need to keep it in exponential form, so no logs. I need also to know how it would be solved if the bases were equal. Thanks.

Answer 1

solving it is essentially getting rid of the exponential

Answer 2

Can you show me how to do that?

Answer 3

you need logs...willing to give it a chance?

Answer 4

But logs is the opposite, or inverse of exponential. Won't that mean that we will be solving for the opposite of what I need?

Answer 5

x = log6/log23

Answer 6

no...you're solving for x. The way to get rid of a power of x is to apply logs on both sides.

Answer 7

Why does this help?

Answer 8

so log23^x = log6

xlog23 = log6

x= log6/log23

Answer 9

Oh, ok. Is this how to solve it or how to solve it if they had equal bases?

Answer 10

yeah this is how you solve it. If they were equal bases, the powers would be equal too, provided that they are the only two terms in the equation

Answer 11

like for example

23^x = 23^4

then x = 4

Answer 12

So, how do i solve x= log6/log23

Answer 13

are you allowed to use a calculator in your exam? If you can, then you are allowed to just use the log button on your calculator. Otherwise you will need log tables which should be available to you at the exam if you are supposed to do these problems

Answer 14

or quite simply, you could leave the answer in exact form by leaving it as log6/log23.

Answer 15

I don't have either. This is not like a huge exam. More like what did you learn? Only, the lessons don't explain it all that well, which is why I sought help. They say I can use any resources, but I don't have a calculator and neither does my PC.

Answer 16

I am doing Edexcel's GCE further maths. You could ask your teachers if you are allowed to use a calculator. If you are, then maybe you should get one if you can.

Answer 17

how do you do your sines and cosines, etc? :/

Answer 18

I want to get a calculator, I just can't right this minute. I can use one, but if you have one, do you think you could calculate it for me? And I don't know what sines and cosines are.

Answer 19

Uh huh...which grade or form are you in? Or which level of maths do u do? If you really need it, log6/log23 is 0.571 to 3dp.

Answer 20

I am not a math person. I don't understand much of Algebra 2, but I have to do it. I am in 9th grade.

Answer 21

there really is no need to decimalize that though...leaving it as log6/log23 is perfectly fine

Answer 22

never heard of "sin" , "cos" "tan" ?

Answer 23

No. Never.

Answer 24

Do you have time to help me with one more?

Answer 25

hmm...maybe that comes later on..I am not from the US so I wouldn't know your syllabus :/ and yeah sure...

Answer 26

Ok. I will post it as a new question so that I can give you another metal. Watch for my username.

Answer 27

haha :) but I don't really care much for medals though

Answer 28

This is my first and last time here...just came on this website for absolutely no reason