Question

Can u simplify without calculators? i think the numbers are huge [question attached]

Answer 1

try expressing the fractions with the same denom. the no.s before logs go as exponents and then all u have to do is multiply them since the base is the same =10
I guess

Answer 2

well expansion isn't a problem, simplification is... i think (16/15)^7 and (25/24)^5 is just too big a number to simplify... i won't even mention the 80 and 81

Answer 3

yea yea...I'm just to lazy to think abt this.....
hmm w8...silly me...all u have to do is multiply and simplify if ucan

Answer 4

\( \log 2 \)

Answer 5

We have
\[ 7\log{\frac{16}{15}}+5\log{\frac{25}{24}}+3\log{\frac{81}{80}}\]
Now
\[ 7\log{\frac{2^4}{3\times 5}}+5\log{\frac{5^{2}}{2^{2} \times 6 }}+3\log{\frac{3^{4}}{2^{4}\times 5}}\]
now we can write this as
\[28 \log 2- 7 \log {3\times 5} +10 \log 5- 5 \log {2^2\times 6}+ 12 \log {3} - 3\log{2^4\times 5} \]
\[28 \log 2- 7 \log {3}-7 \log { 5} +10 \log 5- 10 \log 2-5 \log{ 6}+ 12 \log {3} - 12\log{2}-3\log { 5} \]
Let's combine the terms

\[ 6 \log 2 + 5\log 3- 5\log 6\]

further simplified as
\[ 6 \log 2 + 5\log 3- 5\log 3-5\log 2\]
we get finally
\[ \log 2\]

Answer 6

OMG!! Ash ur awesome!!!!!

Answer 7

Use: \( \log A + \log B = \log (AB) \) and \( \log ( M/N) = \log M - \log N \)

Answer 8

Thanks Angela :)

Answer 9

thanks a ton ash :) that was a pretty comprehensive explanation! :D thanks :)

Answer 10

Welcome Saso:D

Answer 11

it'd difficult when u know the log rules and can't apply them properly, especially when a number can be obtained by the product of many different pairs. thanks ffm too :) I appreciate your contribution :)

Answer 12

i had to use a calculator to get the answer initially :(

Answer 13

Glad to help saso :)