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V125BC [204]
3 years ago
6

Ln a+3 + ln a-3 = ln 16

Mathematics
1 answer:
Lilit [14]3 years ago
7 0
\bf log_{{  a}}(xy)\implies log_{{  a}}(x)+log_{{  a}}(y)\\\\
and\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
ln(a+3)+ln(a-3)=ln(16)\implies ln[(a+3)(a-3)]=ln(16)
\\\\\\
ln[a^2-3^2]=ln(16)\impliedby \textit{removing ln() from both sides}
\\\\\\
a^2-9=16\implies a^2=16+9\implies a^2=25
\\\\\\
a=\pm\sqrt{25}\implies a=\pm 5
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Step-by-step explanation:

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Whats the missing varible for the following questions ​
marta [7]

Answers + Explanation:

1. x + 7 =10

if you know some number + 7 is 10, you know that number must be 3.

so, x = 3

the other way you can solve this would be:

x + 7 = 10

subtract 7 from both sides of the equation

x + 7 - 7 = 10 - 7

x = 3

2. Use the same strategy:

x - 8 = -5

add 8 to both sides of the equation

x - 8 + 8 = -5 + 8

x = 3

3. x + 3 = 6

subtract 3 from both sides of the equation

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4. x + 6 = 3

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7 0
2 years ago
R/16 + 6 = 7 pLEASE HELP WITHIN A HOUR OR TWO
cestrela7 [59]

Solution:

<u>Simplify the equation and solve for r.</u>

  • r/16 + 6 = 7
  • => r/16 = 7 - 6
  • => r/16 = 1
  • => r = 16

The value of r is 16.

<u>Check:</u>

  • r/16 + 6 = 7
  • => 16/16 + 6 = 7
  • => 1 + 6 = 7
  • => 7 = 7 (Proved correct)
7 0
2 years ago
Read 2 more answers
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