Advanced Functions Help!?

Solve for x algebraically: (3^𝑥) + (3^𝑥+1) = (11^𝑥) + (11^𝑥+1) 

Leave your answer as an exact value (do not round).

6 Answers

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  • Philip
    Lv 6
    1 month ago

    3^(x) +3^(x+1) = 11^(x) + 11^(x+1).ie.,;

    4*(3^x) = 12*(11^x);

    3^x = 3(11^x);

    (3/11)^x = 3;

    x = ln3/ln(3/11) or ln3/[ln3 - ln11].

  • sepia
    Lv 7
    1 month ago

    3^x + 3^(x + 1) = 11^x + 11^(x + 1) 

    4 × 3^x = 12 × 11^x

    3^x = 3 × 11^x

    Real solution:

    x ≈ -0.84555

  • ?
    Lv 7
    1 month ago

    (3ˣ) + (3ˣ⁺¹ + 1) = (11ˣ) + (11ˣ⁺¹ + 1)

    3ˣ + 3ˣ•3 + 1 = 11ˣ + 11ˣ•11 + 1

    3ˣ (1+3) + 1 = 11ˣ (1 + 11) + 1

    3ˣ (4) + 1 = 11ˣ (12) + 1

    3ˣ (4) = 11ˣ (12)

    ln [3ˣ (4)] = ln [11ˣ (12)]

    ln 3ˣ + ln 4 = ln 11ˣ + ln 12

    x ln 3 + ln 4 = x ln 11 + ln 12

    x ln 3 - x ln 11 = ln 12 - ln 4

    x (ln 3 - ln 11) = ln 12 - ln 4

    ........ln 12 - ln 4

    x = ----------------

    .......ln  3 - ln 11

    ..........// Note that

    ..........// ln 12 = ln 2²•3 = 2 ln 2 + ln 3

    ..........// ln 4 = ln 2² = 2 ln 2

    ..........// So,

    ..........// ln 12 - ln 4 = 2 ln 2 + ln 3 - 2 ln 2

    ..........// ln 12 - ln 4 = ln 3

    ............ln 3

    x = ----------------

    .......ln  3 - ln 11

    ..........// ln 3 - ln 11 = ln (3/11)

    ..........ln 3

    x = -------------....................ANS

    .......ln  (3/11)

  • rotchm
    Lv 7
    1 month ago

    Assuming you meant:  3^𝑥 + 3^(𝑥+1) = 11^𝑥 + 11^(𝑥+1)  

    3^𝑥 + (3^𝑥)(3^1) = 11^𝑥 + (11^𝑥)(11^1)

    3^𝑥 + 3(3^𝑥) = 11^𝑥 +11(11^𝑥)

    4(3^𝑥) = 12(11^𝑥)

    3^𝑥 = 3(11^𝑥)

    3^𝑥 / 11^𝑥= 3

    (3/11)^𝑥 = 3.  take the Log each side, no matter the base.

    x L(3/11) = L 3, thus x = L3 / L(3/11) =  L3 / (L3 - L11), no matter the base. 

    If you take base 3, this becomes x = 1/(1 - L₃11) which is a fun way to write the answer. Or just L3 / L(3/11)  in any base; so you can use log_10 or Ln etc and you still get the same (numerical) answer. 

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  • Amy
    Lv 7
    1 month ago

    Do you perhaps have the parentheses in the wrong place? They do nothing at all where they are now.

    3^x + 3^x + 1 = 11^x + 11^x + 1

    2 * 3^x = 2 * 11^x

    (3/11)^x = 1

    x = 0

    A more interesting problem would have been 

    3^x + 3^(x+1) = 11^x + 11^(x+1)

    (1 + 3) * 3^x = (1 + 11) * 11^x

    3^x = 3 * 11^x

    (3/11)^x = 3

    x = log(3) /  log(3/11) 

  • 1 month ago

     (3^𝑥) + (3^𝑥+1) = (11^𝑥) + (11^𝑥+1) 

    following the rules of precedence, not guessing at your use of parans, this is

     (3^𝑥) + (3^𝑥+1) = (11^𝑥) + (11^𝑥+1) 

     (3^𝑥) + (3^𝑥) +1 = (11^𝑥) + (11^𝑥) +1

    2•(3^𝑥) = 2•(11^𝑥)

    3^𝑥 = 11^𝑥

    log both sides

    x log3 = x log11

    x=0 is only solution

    BUT, it may be (expressed incorrectly) :

    3^𝑥 + 3^(𝑥+1) = 11^𝑥 + 11^(𝑥+1) 

    3^𝑥 + (3^𝑥)(3^1) = 11^𝑥 + (11^𝑥)(11^1) 

    3^𝑥 + 3(3^𝑥) = 11^𝑥 +11(11^𝑥)

    4(3^𝑥) = 12(11^𝑥)

    3^𝑥 = 3(11^𝑥)

    log both sides

    x log 3 = log 3 + x log 11

    x(log3 – log11) = log3

    x = log3 / (log3 – log11)

    many different variations of this with log of various bases, but all are the same number

    eg

    x log₃3 = log₃3 + x log₃11

    x = 1 + x log₃11

    x(1 – log₃11) = 1

    x = 1 / (1 – log₃11)

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