What is ln(49)*log_{7}(e^2) ?

The solution to ln(49)*log_{7}(e^2) is 4

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simplify ln(49)*log_{7}(e^2)

simplify

ln (49) \cdot lo g_{7} (e^{2})

4

Solution steps

ln (49) lo g_{7} (e^{2})

Simplify

lo g_{7} (e^{2}) : 2 lo g_{7} (e)

= 2 ln (49) lo g_{7} (e)

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

ln (49) lo g_{7} (e) = \frac{ln ( 49 )}{ln ( e )} \cdot \frac{ln ( e )}{ln ( 7 )}

= 2 \cdot \frac{ln ( 49 )}{ln ( e )} \cdot \frac{ln ( e )}{ln ( 7 )}

Refine

= 2 \cdot \frac{ln ( 49 )}{ln ( 7 )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{ln ( 49 ) \cdot 2}{ln ( 7 )}

Simplify

ln (49) : 2 ln (7)

= \frac{2 \cdot 2 ln ( 7 )}{ln ( 7 )}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 ln ( 7 )}{ln ( 7 )}

Cancel the common factor:

ln (7)

= 4

(\frac{2}{4})^{4}

(12 y^{2} + 17 y - 4) + (9 y^{2} - 13 y + 3)

(3 x - 9)^{2}

\frac{2 x - 4 y}{x ^{2} - 2 x y}