Question #03bcb

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Question #03bcb

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Answer 1 · 2017-03-28T02:46:56

$a = \frac{2 ln 5 + ln 2}{5 ln 5 - 2 ln 2}$ $a=(2ln5+ln2)/(5ln5-2ln2)$

Explanation:

Take the natural logarithm of both sides.
$ln (5^{5 a - 2}) = ln (2^{2 a + 1})$ $ln(5^(5a-2))=ln(2^(2a+1))$
Use the log law $log (a^{b}) = b log (a)$ $log(a^b)=blog(a)$ to make the index a coefficient.
$(5 a - 2) ln (5) = (2 a + 1) ln (2)$ $(5a-2)ln(5)=(2a+1)ln(2)$
Expand Brackets
$5 a ln (5) - 2 ln (5) = 2 a ln (2) + ln (2)$ $5aln(5)-2ln(5)=2aln(2)+ln(2)$
Combine like terms on either side of equality
$5 a ln (5) - 2 a ln (2) = 2 ln (5) + ln (2)$ $5aln(5)-2aln(2)=2ln(5)+ln(2)$
Factorise by a on the left hand side
$a [5 ln (5) - 2 ln (2)] = 2 ln (5) + ln (2)$ $a[5ln(5)-2ln(2)]=2ln(5)+ln(2)$
Divide both sides by $5 ln (5) - 2 ln (2)$ $5ln(5)-2ln(2)$
$\frac{a [5 ln (5) - 2 ln (2)]}{5 ln (5) - 2 ln (2)} = \frac{2 ln (5) + ln (2)}{5 ln (5) - 2 ln (2)}$ $(a[5ln(5)-2ln(2)])/(5ln(5)-2ln(2))=(2ln(5)+ln(2))/(5ln(5)-2ln(2))$
Cancelling on left hand side leaves;
$a = \frac{2 ln 5 + ln 2}{5 ln 5 - 2 ln 2}$ $a=(2ln5+ln2)/(5ln5-2ln2)$

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Question #03bcb

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