3^(x+4)=2^(1-3x)
Take logarithm on both sides and use the exponent rule of logarithms:
log(ab) = b*log(a)
So applying the rule to the given equation:
3^(x+4)=2^(1-3x)
log(3^(x+4)) = log(2^(1-3x))
(x+4)log(3) = (1-3x)log(2)
Solve the resulting linear equation after substituting numerical values of log(3) and log(2). You can use either natural log (to the base e) or common log (to the base 10).
I get x=-1.2 approximately.
i'm sorry i don't understand the last part :(
Evaluate log(3) and log(2) using logarithm to the base 10.
log103 = 0.4771
log102 = 0.3010
(x+4)log(3) = (1-3x)log(2)
0.4771(x+4) = 0.3010(1-3x)
Solving for x, I get x=1.16 approx.
If you had used logarithm to the base e (natural log), you would have got the same answer of x=1.16 approx.
To solve the equation 3^(x+4) = 2^(1-3x), we need to get rid of the exponents. We can do this by taking the logarithm of both sides of the equation.
1. Begin by taking the logarithm (any base) of both sides of the equation. Let's use the natural logarithm (ln) for simplicity:
ln(3^(x+4)) = ln(2^(1-3x))
2. Use the logarithm property: ln(a^b) = b * ln(a), to simplify the equation:
(x+4) * ln(3) = (1-3x) * ln(2)
3. Expand and rearrange the equation:
x * ln(3) + 4 * ln(3) = ln(2) - 3x * ln(2)
4. Group the x terms together and the constant terms together:
x * (ln(3) + 3ln(2)) = ln(2) - 4 * ln(3)
5. Divide both sides of the equation by (ln(3) + 3ln(2)):
x = (ln(2) - 4 * ln(3)) / (ln(3) + 3ln(2))
6. Use a calculator to evaluate the expression on the right side. Remember to use parentheses accordingly:
x ≈ -0.872
So the solution to the equation 3^(x+4) = 2^(1-3x) is approximately x = -0.872.