Question

Chapter 5:Indices and Logarithms
@ganeshie8

Answer 1

solve the following equation

Answer 2

\[3\log _{2}5+\log _{2}(2x-1)-\log _{2}(x+2)=1\]

Answer 3

Notice
\(3\log_2 (5) = \log_2 (5^3)\)

Answer 4

write the left hand stuff as a single log

Answer 5

okay

Answer 6

\[\log _{2}5^3+\log _{2}2+\log _{2}x+\log _{2}-1-\log _{2}x+\log _{2}2=1\]

Answer 7

Hey no

Answer 8

below is false :
\[\log_b (x+y) \ne \log_b(x) + \log_b(y)\]

Answer 9

Then factorize?

Answer 10

\[\log _{2}(5^3+2x-1-x-2)=1\]

Answer 11

\[3\log _{2}5+\log _{2}(2x-1)-\log _{2}(x+2)=1\]

\[\log _{2}5^3+\log _{2}(2x-1)-\log _{2}(x+2)=1\]

combine the first two logs using product property:
\[\log _{2} \left(5^3(2x-1)\right)-\log _{2}(x+2)=1\]

Answer 12

combine the logs again using quotient property of logarithm :
\[\log_2\left(\dfrac{5^3(2x-1)}{x+2}\right)=1\]

Answer 13

right ?

Answer 14

\[\frac{ 5^3(2x-1) }{ x+2 }=2\]

Answer 15

yes!

Answer 16

cross multiply and solve x

Answer 17

\[5^3(2x-1)=2(x+2)\]

Answer 18

\[125(2x-1)=2x+4\]

Answer 19

\[250x-125=2x+4\]

Answer 20

\[248x=125+4\]

Answer 21

\[x=0.5202\]

Answer 22

Thnx @ganeshie8 :)

Answer 23

good job !