Solve the following exponential equation. 3^x^-^3=5^4^x^+^1 The answer is [tex]\frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}[/tex

Question

Solve the following exponential equation.
3^x^-^3=5^4^x^+^1
The answer is \frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}. I was given the answer, but I do not know the steps to get to it.

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Jezebel 4 years 2021-08-17T19:23:54+00:00 1 Answers 12 views 0

Answers ( )

  1. linhdan
    0
    2021-08-17T19:24:57+00:00
    Reply

    Answer:

    x = \frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}

    Step-by-step explanation:

    1) Since we cannot cancel out the bases, take the natural log of both sides.

    3^x^-^3 = 5^4^x^+^1\\ln 3^x^-^3 = ln 5^4^x^+^1

    2) Now, use the power property of logarithms to move the exponents out in front like so:

    ln 3^x^-^3 = ln 5^4^x^+^1\\(x-3)ln3 = (4x + 1) ln5

    3) Now, expand out the terms inside the parentheses:

    (x-3)ln3 = (4x + 1) ln5\\xln3-3ln3=4xln5+ln5

    4) Now, move all the terms with the x variable onto one side:

    xln3-3ln3=4xln5+ln5\\xln3-4xln5 = 3ln3+ln5

    4) Now, factor the x out of the x terms on the left side:

    xln3-4xln5 = 3ln3+ln5\\x(ln3-4ln5)=3ln3+ln5

    5) Finally, divide both sides by ln3-4ln5 to isolate and solve for x.

    x(ln3-4ln5)=3ln3+ln5\\x = \frac{3ln3+ln5}{ln3-4ln5}

    Thus, x = \frac{3ln(3)+ln(5)}{ln(3)-4ln(5)}.

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