Question

Chapter 5:Indices and Logarithms
@ganeshie8

Answer 1

\[Given~that~\log _{2}x-\log _{x}8=2,find~the~values~of~x.\]

Answer 2

Hint : \[\log_y x = \dfrac{1}{\log_x y}\]

Answer 3

\[=\frac{ 1 }{ \log _{2}x }\]

Answer 4

\[\log_2 x - \log_x 8 = 2\]

\[\log_2 x - \log_x 2^3 = 2\]

\[\log_2 x - 3\log_x 2 = 2\]

\[\log_2 x - 3\dfrac{1}{\log_2 x} = 2\]

Answer 5

let \(t = \log_2 x\) :
\[t - 3\frac{1}{t} = 2\]

Answer 6

turn it into a quadratic and solve \(t\)

Answer 7

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

Answer 8

nope

\[t - 3\frac{1}{t} = 2 \]

multiply \(t\) throught out :
\[t^2 - 3 = 2t \]

which is same as
\[t^2-2t - 3 = 0 \]

see if you can factor this and solve \(t\)

Answer 9

t=3 and t=-2

Answer 10

try again

Answer 11

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

Answer 12

heyy first solve the quadratic

Answer 13

t=3 and t=-2 are wrong

Answer 14

\[t^2-2t-3=0\]\[t(t-2)-3=0\]\[t=3~and~t=2\]

Answer 15

\[t^2 - 2t -3 = 0\]

\[t^2 - 3t+t -3 = 0\]

\[t(t - 3)+1(t -3) = 0\]
\[(t - 3)(t+1) = 0\]

\[t = 3, -1\]

Answer 16

okay :)

Answer 17

and previously we let \(\log_2 x=t \)

so
\[\log_2 x = 3 \implies x = 2^3\]
\[\log_2 x = -1 \implies x = \frac{1}{2}\]

Answer 18

x=8,x=1/2

Answer 19

Thnx @ganeshie8 :)

Answer 20

yw :)