Question

help please

solve each equation. Give an exact solution
log(2x-1)+logx=1

Answer 1

@ganeshie8

Answer 2

@ganeshie8 log(2x-1)(x)=1 right ???

Answer 3

correct !

Answer 4

did u notice, you're not given any base for logs ?

Answer 5

wat does that mean ?

Answer 6

so its log2x^2-2=1 the base is 10

Answer 7

my that should be a -x

Answer 8

2x^2-x=1

Answer 9

yes base is 10

Answer 10

\( \log(2x-1)(x)=1 \)

is same as

\( \log_{10}(2x-1)(x)=1 \)

Answer 11

next, use the log property :

\(\large \log_a b = c \iff b = a^c\)

Answer 12

\( \log_{10}(2x-1)(x)=1 \)

\( (2x-1)(x)=10^1 \)

\( 2x^2-x=10 \)

Answer 13

solve \(x\)

Answer 14

thats where i get stuck

Answer 15

knw how to factor a quadratic ?

Answer 16

\( 2x^2-x=10 \)

\( 2x^2-x-10 = 0 \)

Answer 17

its a quadratic equation,
and we can factor it

Answer 18

or we can use quadratic formula

Answer 19

which one u familiar wid ?

Answer 20

formula

Answer 21

use it

Answer 22

first figure out below values :
\(a = ?\)
\(b = ?\)
\(c = ?\)

Answer 23

formula :

\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Answer 24

\(2x^2 - x - 10 = 0\)

compare this wid :

\(ax^2 + bx + c = 0\)

Answer 25

what are a, b, c values ?

Answer 26

i got x=5 or x=-4

Answer 27

try again

Answer 28

\(\large x = \frac{--1 \pm \sqrt{1-4(2)(-10)}}{2(2)}\)

\(\large x = \frac{1 \pm \sqrt{81}}{4}\)

\(\large x = \frac{1 \pm 9}{4}\)

\(\large x = \frac{1 + 9}{4}\) , \(\large x = \frac{1 - 9}{4}\)

\(\large x = \frac{10}{4}\) , \(\large x = \frac{-8}{4}\)

\(\large x = \frac{5}{2}\) , \(\large x = -2\)

Answer 29

however since log cannot suck in negative values,
we need to throw away x = -2 value

Answer 30

therefore, the only solution for given equation is :

\(\large x = \frac{5}{2}\)

Answer 31

see if that makes more or less sense...