# 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

- PhilipLv 61 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].

- sepiaLv 71 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 71 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)

- rotchmLv 71 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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- AmyLv 71 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)

- billrussell42Lv 71 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)