3 years ago

Given: log2 = a, log7 = b. Find: log56

1 answer:

7 0

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \end{array} ~\hfill \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ log_a\left( x^b \right)\implies b\cdot log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf log(56)~~ \begin{cases} 56=2\cdot 2\cdot 2\cdot 7\\ \qquad 2^2\cdot 2\cdot 7 \end{cases}\implies log(2^2\cdot 2\cdot 7) \\\\\\ log(2^2)+\stackrel{a}{log(2)}+\stackrel{b}{log(7)}\implies 2\stackrel{a}{log(2)}+a+b\implies 2a+a+b\implies 3a+b$

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Read 2 more answers

Solve the equation 1/3m+6m-9/3m=3m-3/4m

vekshin1

M
=
0

O
r

m
=
23
15

Explanation:
1
3
m
+
6
m
−
9
3
m
=
3
m
−
3
4
m

⇒
1
+
6
m
−
9
3
m
=
3
m
−
3
4
m

⇒
−
8
+
6
m
3
m
=
3
m
−
3
4
m

⇒
(
−
8
+
6
m
)
×
4
m
=
(
3
m
−
3
)
×
3
m

⇒
−
32
m
+
24
m
2
=
9
m
2
−
9
m

⇒
−
32
m
+
24
m
2
−
9
m
2
+
9
m
=
0

⇒
15
m
2
−
23
m
=
0

⇒
m
(
15
m
−
23
)
=
0

m
=
0

Or

15
m
−
23
=
0
⇒
15
m
=
23
⇒
m
=
23
15

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