Use the following equation: log2x + log2(x – 7) = 3 The answer to the function would be: Select one: a. 8 b. –1 c. 8 or –1 d. –8 or 1
You can use here: \[\log(a) + \log(b) = \log(a \times b)\]
well here is a tip... you can't find the log of a negative number.... so that should help identify the answer
is 2 in base @chantz417
Yes sorry just got back
where 2 is in base of logarithm
u there @chantz417
lol... how can you substitute x = -1 what is the value of \[\log_{2}(-1) =..?\]
wow... understand some basics of logarithms
you can after u make it log(x(x-7)) after that u get x(x-7) > 0 implying x can be either greater then 7 or less than -1 read logarithms and functions carefully @campbell_st
u can just put x=-1 in logx that's wrong but you can after u make x(x-7) as it will come out positive @campbell_st
can't* not can typed wrongly
have a look at the original equation you get \[\log_{2}( -1) + \log(-1 - 7) = 3\] tell me how that works...
the only solution is 8....
seperately it doesn't but if u make it something like this log(-1)(-1-7)
so i missed u clicked u are right we must see it other perspective as well @campbell_st gud luck !!!
just test the solution in the original equation... the only solution that works is x = 8
ans 8 only
you are provided with the answer choices.... the easy solution with out calculation or algebraic manipulation is substitution... so substitute the choices and see what you get
not every one is that intelligent @campbell_st you need to xplain others
u and i can do that but not every one
well I'm in high school... and have been taught to always check the solutions make sense.
it's ohk but doesn't always work that way @campbell_st still u have a long way to go
u will get options sometimes u can't just check them out
lol... why make things hard.... use the obvious... another good problem solving skill.
u'll learn why to solve ... later it might look that easy today .... @campbell_st
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