Question

Chapter 5:Indices and Logarithms
@ganeshie8

Answer 1

\[\frac{ 49^{2n+1}\times14^{6-2n} }{ 56^{2n-5} }\]

Answer 2

Simplify the following

Answer 3

@hartnn

Answer 4

start by factoring 14
and 56

Answer 5

\[==\frac{ 49^{2n+1}\times(7\times2)^{6-2n} }{ (7\times8)^{2n-5} }\]

Answer 6

Do it with 49 as well. You may want to factor all bases down to primes...

Answer 7

okay

Answer 8

\[=\frac{ (7\times7)^{2n+1}\times(7\times2)^{6-2n} }{ (7\times8)^{2n-5} }\]

Answer 9

\[=\frac{ (7\times7)^{2n+1}\times(7\times2)^{6-2n} }{ (7\times2\times2\times2)^{2n-5} }\]

Answer 10

\[=\frac{ 7^{2n+1}\times7^{2n+1}\times7^{6-2n}\times2^{6-2n} }{ 7^{2n-5}\times2^{2n-5}\times2^{2n-5}\times2^{2n-5} }\]

Answer 11

\[=\frac{ 7^{2n+1+2n+1+6-2n}\times2^{6-2n} }{ 7^{2n-5}\times2^{8n-15} }\]

Answer 12

\[=\frac{ 7^{2n+8}\times2^{6-2n} }{ 7^{2n-5}\times2^{8n-15} }\]

Answer 13

\[=7^{2n+8}\times2^{6-2n}\div7^{2n-5}\times2^{8n-15}\]

Answer 14

\[=7^{2n+8-2n+5}\times2^{6-2n-8n+15}\]

Answer 15

\[=7^{13}\times2^{21-10n}\]

Answer 16

@campbell_st

Answer 17

I got \[=7^{13}\times2^{21-10n}\] but the answer in the book say it's \[=7^{13}\times2^{21-8n}\]

Answer 18

Am i correct?

Answer 19

book is right

Answer 20

\[\large{
\frac{ 49^{2n+1}\times14^{6-2n} }{ 56^{2n-5} }\\\\
=\frac{7^{2(2n+1)}\times7^{6-2n}\times2^{6-2n}}{7^{2n-5}\times8^{2n-5}}\\\\
=\frac{7^{4n+2+6-2n}\times2^{6-2n}}{7^{2n-5}\times2^{3(2n-5)}}\\\\
=\frac{7^{2n+8}\times2^{6-2n}}{7^{2n-5}\times2^{6n-15)}}\\\\
=7^{2n+8-2n+5}\times2^{6-2n-6n+15}\\
=7^{13}\times 2^{21-8n}
}\]

Answer 21

an \(\color{red}{\text e}\)rror in your working
\[=\frac{ 7^{2n+1}\times7^{2n+1}\times7^{6-2n}\times2^{6-2n} }{ 7^{2n-5}\times2^{2n-5}\times2^{2n-5}\times2^{2n-5} }\]\[=\frac{ 7^{2n+1+2n+1+6-2n}\times2^{6-2n} }{ 7^{2n-5}\times2^{\color{red}6n-15} }\]

Answer 22

okay know i get :)
yay

Answer 23

Thnx @UnkleRhaukus