Question

7log16/15+ 5log25/24+ 3log81/80

find it's value

Answer 1

Hey...

Refer to attachment

Hope it helps ^_^

Answer 2

log 2 is the answer for $\bold{7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right)}$

Given:

$7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right)$

To find:

The value of $7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right) = ?$

Solution :

Simplifying the question we get:

$7(\log 16-\log 15)+5(\log 25-\log 24)+3(\log 81-\log 80)$

Simplifying those number which can be used as square numbers like 16, 25, 81

$7\left(\log 4^{2}-\log 15\right)+5\left(\log 5^{2}-\log 24\right)+3\left(\log 9^{2}-\log 80\right)$

Placing the numbers we get:

$7\left(\log 4^{2}-\log (3 \times 5)\right)+5\left(\log 5^{2}-\log (6 \times 4)\right)+3\left(\log 9^{2}-\log (8 \times 10)\right)$

Simplifying the log values, after simplifying changing the values of log4 and log9 to 2log2 and 2log3.

$7(2 \log 4-\log 3+\log 5)+5(2 \log 5-\log 6+\log 4)+3(2 \log 9-\log 8+\log 10)$

Adding all the similar log value together we get:

$28 \log 2-15 \log 2-12 \log 2-7 \log 3+5 \log 3+12 \log 3-7 \log 5+10 \log 5-3 \log 5$

$\begin{array}{l}{28 \log 2-27 \log 2-12 \log 3+12 \log 3-10 \log 5+10 \log 5} \\ {=\log 2}\end{array}$

Therefore, after solving the values of log we find the value of the equation that is log2.