Question

logy+log(y+21)=2

Answer 1

\[

\log(y(y+21)) = 2
\]

Answer 2

y^2+21y-2

Answer 3

No, do you know the base of this logarithm?

Answer 4

10

Answer 5

Then you want to do \[
10^{\log(y(y+21))}=10^2
\]

Answer 6

so it would be?

Answer 7

well, you get \[
y^2+21y = 100
\]

Answer 8

Then you can solve it normally.

Answer 9

how @wio

Answer 10

It's a quadratic equation. You got this.

Answer 11

rearrange to

y^2+21x-100=0

then you could apply quadratic formula,

or simply
look for factors of -100, that add to 21

Answer 12

what's the factor?

Answer 13

y^2+21y-100=0 * typo

Answer 14

factors of -100, are numbers that multiply together to equal -100

Answer 15

There is no way they would expect you to solve this problem if you didn't already know how to solve a quadratic equation.

Answer 16

i dont get this

Answer 17

for example 2 and -50 are factors of -100,
these factors add to -48

but you want factors that add to give 21 (not -48)

Answer 18

what are two other numbers that when multiplied together give -100?
@shay928

Answer 19

50 2

Answer 20

25 -4

Answer 21

right, 25 and -4 multiply to give -100, and add to give 21, good work

Answer 22

x=25, y=-4

Answer 23

so you can factorise

y^2+21y-100=0
(y+25)(y-4) =0

Answer 24

now you have a product that equals zero
for this to be true
one of the terms in brackets must equal zero

y+25=0 or y-4=0

Answer 25

solve these equations separately, to get the two solutions for y

Answer 26

-25 4

Answer 27

yep

Answer 28

thanks

Answer 29

but if you check these in the original equation, one of them wont fit,

only the log of a positive number will be real

Answer 30

so through away one of these solution to the quadratic, and you will be left with

y= ....

Answer 31

25

Answer 32

nope, -25 was a solution to the quadratic,

but
log(-25)+log(-25+21)=2

wont work

Answer 33

4

Answer 34

if we check y=4 in the original equation

log(4)+log(4+21)=
log(4(25))=
log(100)=2

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