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Carsmypets222
Carsmypets222 asked in Science & MathematicsMathematics · 9 years ago

Logarithm Problems...?

I. 6^ (3x+5) = 1

II. log(base: 8) (2x+1) = -1

i dont know how to these... and can you please include steps, so i can understand how to do it.

thank you! :)

Update:

the answers were given: I. -5/3

II. -7/16

4 Answers

Relevance
  • The Duke of Snowdon
    Lv 5
    9 years ago
    Favorite Answer

    I. For 6 raised to any power to be equal to 1, the exponent must be zero!

    Hence, 3x + 5 = 0

    x = -5/3 or -1 2/3

    II. If log(base: 8) (2x+1) = -1 that means raising 8 to the power -1 gives 2x + 1

    so 1/8 = 2x + 1

    hence 16x + 8 = 1 or 16x = -7

    So x = -7/16

  • Mike G
    Lv 7
    9 years ago

    i) Recall a^0 = 1

    3x+5 = 0

    x = -5/3

    ii) 2x+1 = 8^-1 = 1/8

    2x = -7/8

    x = -7/16

  • Iggy Rocko
    Lv 7
    9 years ago

    I. 6^ (3x+5) = 1

    log(base 6)(6^(3x+5)) = log(base 6)1

    3x + 5 = 0

    3x = -5

    x =-5/3

    II. log(base: 8) (2x+1) = -1

    8^-1 = 2x + 1

    1/8 = 2x + 1

    -7/8 = 2x

    x = -7/16

  • JOS J
    Lv 7
    9 years ago

    1.

    x = - 5/3

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