Question
(2*log7 16)/(log3(√10+1)+log3(√10-1)log7 2
Answers
Steve
I suspect a typo, since
log3(√10+1)+log3(√10+1) = log3(10-1) = log3(9) = 2
the log7 of powers of 2 is not amenable to simplification.
Is there a typo? At the very least, the parentheses are not balanced.
log3(√10+1)+log3(√10+1) = log3(10-1) = log3(9) = 2
the log7 of powers of 2 is not amenable to simplification.
Is there a typo? At the very least, the parentheses are not balanced.
MathMate
Fractions have implicit parenthese around the denominator and numerator.
I have a feeling that the two log<sub>3</sub> terms both belong to the denominator, in which case the expression should have been written, according to the BEDMAS rules,
(2*log<sub>7</sub> 16)/<b>(</b>(log<sub>3</sub>(√10+1)+log<sub>3</sub>(√10-1)log<sub>7</sub> 2 <b>)</b>
in which case:
(2*log<sub>7</sub>2<sup>4</sup>)/<b>(</b>(log<sub>3</sub>(√10+1)(√10-1) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>(log<sub>3</sub>(10-1) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>(log<sub>3</sub>(9) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>2 log<sub>7</sub> 2 <b>)</b>
=4
I have a feeling that the two log<sub>3</sub> terms both belong to the denominator, in which case the expression should have been written, according to the BEDMAS rules,
(2*log<sub>7</sub> 16)/<b>(</b>(log<sub>3</sub>(√10+1)+log<sub>3</sub>(√10-1)log<sub>7</sub> 2 <b>)</b>
in which case:
(2*log<sub>7</sub>2<sup>4</sup>)/<b>(</b>(log<sub>3</sub>(√10+1)(√10-1) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>(log<sub>3</sub>(10-1) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>(log<sub>3</sub>(9) log<sub>7</sub> 2 <b>)</b>
=(8*log<sub>7</sub>2)/<b>(</b>2 log<sub>7</sub> 2 <b>)</b>
=4
MathMate
Just a lucky guess, but you were right in the first place!