Question

Help please ill give a medal (:

Answer 1

What is the solution of the equation \[4^{x+1}=21\]?

Answer 2

do u know logarihms ?

Answer 3

*logarithms

Answer 4

no

Answer 5

then which topic does this question belong to ?

Answer 6

logarithms

Answer 7

im just not sure how to do them because its review

Answer 8

oh, okk..
first taking log on both sides, we get
\(\log(4^{x+1})=\log 21\)

Answer 9

and then what should i do?

Answer 10

then we use the log property that \(\large \log a^b=b\log a\)
so, what about \(\log 4^{x+1}\)
??

Answer 11

ummm i dont understand what to do after that

Answer 12

got this ?