Checking my work: for log2x=log3+log(x+4), i get x=7

[charlesjoy]

good afternoon. Can you please cehck my equation to see if I did it correctly?

log 2x=log 3 +log(x+4)


log2x=log 3 +logx +4 log

log 2x=7 log +1 logx

1log x = 7 log

answer is 7

 

[Deleted member 4993]

Re: Checking my work

good afternoon. Can you please cehck my equation to see if I did it correctly?

log 2x=log 3 +log(x+4)


log2x=log 3 +logx +4 log <<< Incorrect

use

\(\displaystyle log(a\cdot b) \, = \, log(a) \, + \, log(b)\)

log 2x=7 log +1 logx

1log x = 7 log

answer is 7

You could check your answer by putting it back into the given equation

log 2x=log 3 +log(x+4)

log(14) = log 3 + log(11)

log(14) = log (3*11) <<< This does not satify the equality - hence your answer is incorrect.

However, this problem is little complicated - it does not have a solution.

 

[DrMike]

Re: Checking my work

good afternoon. Can you please cehck my equation to see if I did it correctly?

log 2x=log 3 +log(x+4)

Are you very sure you typed this in correctly?