solve with logs
3^(1-2x)=4^(x)
How do I do this also: write the expression as a single logarithm
21 log (base 3)time the cube root of x + log (base 3) * 9x^2 - log (base 5)25
As you have it printed, I would do the following.
(1-2x)log 3 = xlog 4
Solve that. I get x = 0.306
To solve the equation 3^(1-2x) = 4^x using logarithms, we can take the logarithm of both sides of the equation. Since the bases of the exponents are different, we need to choose a common logarithmic base. In this case, we can choose either the natural logarithm (ln) or the common logarithm (log base 10).
Let's use the natural logarithm, so we have:
ln(3^(1-2x)) = ln(4^x)
Using the logarithm rule, we can bring down the exponents:
(1-2x)ln(3) = xln(4)
Next, we can distribute the ln(3) to both terms on the left side:
ln(3) - 2xln(3) = xln(4)
Now, isolate the terms with x on one side and the constant terms on the other side:
ln(3) = xln(4) + 2xln(3)
To simplify, we can factor out an x:
ln(3) = x(ln(4) + 2ln(3))
Finally, divide both sides by (ln(4) + 2ln(3)) to solve for x:
x = ln(3) / (ln(4) + 2ln(3))
This is the solution to the equation using logarithms.