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daser333 [38]
2 years ago
7

Solve. Log8 (6x) = Log8 2 + Log8 (x-4)

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
7 0

Write both sides as powers of 8, then simplify:

8^{\log_86x}=8^{\log_82+\log_8(x-4)}=8^{\log_82}8^{\log_8(x-4)}

\implies6x=2(x-4)

\implies6x=2x-8

\implies4x=-8

\implies x=-2

However, on the left hand side, this would mean we'd have, \log_8-12, which is undefined (assuming the real-valued logarithm), so this equation has no real solution.

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C. sides of 3, 4 and 7

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Step-by-step explanation:

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3 years ago
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egoroff_w [7]

Answer:

Step-by-step explanation:

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Let P = 5, |Q| = 6, R =7 |S| = 2

Note that since Q and S are in modulus sign, they can return both positive and negative values.

P+Q = 5 + 6 (note that the positive value of Q is used since we need the greatest value of P+Q)

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For the least value of P+Q, we will use the negative value of Q as shown

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Similarly:

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Hence the least value of P+Q is 2

NOTE THAT THIS ARE ASSUMED VALUES. ALL YOU NEED IS TO PLUG IN THE VALUES OF P, Q and R THAT YOU HAVE IN CASE THE VALUES DIFFERS.

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3 years ago
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