Five more than three-fourths a number is -19

Mathematics

1 answer:

gogolik [260]2 years ago

3 0

Answer:

5 + 3/4x = -19

Step-by-step explanation:

more than means adding

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Step-by-step explanation:

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Can someone help me in #66

VladimirAG [237]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
$

$\bf log_4\left( \cfrac{\sqrt{x^5y^7}}{zw^4} \right)\implies log_4(\sqrt{x^5y^7})-log_4(zw^4)
\\\\\\
log_4\left[(x^5y^7)^{^\frac{1}{2}}\right]-log_4(zw^4)\implies 
\cfrac{1}{2}log_4\left[(x^5y^7)\right]-log_4(zw^4)
\\\\\\
\cfrac{1}{2}\left[ log_4(x^5)+log_4(y^7) \right] -[log_4(z)+log_4(w^4)]
\\\\\\
\cfrac{1}{2}log_4(x^5)+\cfrac{1}{2}log_4(y^7)-[log_4(z)+log_4(w^4)]$

$\bf \cfrac{1}{2}\cdot 5log_4(x)+\cfrac{1}{2}\cdot 7log_4(y)-log_4(z)-4log_4(w)
\\\\\\
\cfrac{5}{2}log_4(x)+\cfrac{7}{2}log_4(y)-log_4(z)-4log_4(w)$

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