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Agata [3.3K]
3 years ago
10

NEED HELP What is the real value of x in the equation log.2.24 - log.2.3 = log.5.x ?

Mathematics
2 answers:
tia_tia [17]3 years ago
8 0

Answer:

125

Step-by-step explanation:

log2 (24) - log2 (3) = log5 (x)

We know that log a(b) - log a(c) = log a( b/c)

log2 (24/3)  = log5 (x)

log2 (8) = log5 (x)

Rewriting 8 as 2^3

log2 (2^3) = log5 (x)

We know that log a ( a^b) =  b loga (a) and loga (a) =1

3 log2 (2) = log5 (x)

3 *1 =  log5 (x)

3 = log5 (x)

We know log a (b) =c  can be written as a^c =b

5^3 = x

125 =x

sergey [27]3 years ago
6 0

Answer:

x = 125

Step-by-step explanation:

log2(24) - log2(3) = log5(x)

log2(8×3) - log2(3) = log5(x)

log2(8) + log2(3) - log2(3) = log5(x)

log(8) = log5(x)

log2(2³) = log5(x)

3log2(2) = log5(x)

3 = log5(x)

x = 5³

x = 125

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Answer:

  • 171/40 or 4 11/40

Step-by-step explanation:

<h3>AP given</h3>
  • a + b, a - b,  ab, a/b
<h3>To find</h3>
  • 6th term
<h3>Solution</h3>

Common difference

<u>Difference of first two</u>

  • d = (a -b) - (a + b) = -2b

<u>Difference of second two</u>

  • d= ab - (a - b)

<u>Difference of last two</u>

  • d = a/b - ab

<u>Now comparing d:</u>

  • -2b = ab - (a - b)
  • ab - a = - 3b
  • a(1 - b) = 3b
  • a = 3b/(1 - b)

and

  • a/b - ab = -2b
  • a(1/b - b) = -2b
  • a = 2b²/(b² - 1)

<u>Eliminating a:</u>

  • 2b²/(b² - 1) = 3b/(1 - b)
  • 2b/(b+1) = -3
  • 2b = -3b - 3
  • 5b = - 3
  • b = -3/5

<u>Finding a:</u>

  • a = 3b/(1 - b) =
  • 3*(-3/5) *1/(1 - (-3/5)) =
  • -9/5*5/8 =
  • -9/8

<u>So the first term is:</u>

  • a + b = -3/5 - 9/8 = -24/40 - 45/40 = - 69/40

<u>Common difference:</u>

  • d = -2b = -2(-3/5) = 6/5

<u>The 6th term:</u>

  • a₆ = a₁ + 5d =
  • -69/40 + 5*6/5 =
  • -69/40 + 240/40 =
  • 171/40 = 4 11/40
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