How would you solve this for x using logarithims? I tried but when I plugged my answer for x back into the equation it didn't work at all.
4^2x = 9^x-1
Take the log of each side:
2x/ log(4) = (x-1)/log (9)
solve for x.
That's how I started but then I got confused, but I think I get it now thanks.
To solve the equation 4^(2x) = 9^(x-1) using logarithms, follow these steps:
1. Take the logarithm of both sides of the equation. You can use either the common logarithm (log, base 10) or the natural logarithm (ln, base e). In this case, let's use the natural logarithm:
ln(4^(2x)) = ln(9^(x-1))
2. Apply the power property of logarithms, which states that the logarithm of an exponent can be brought down as a coefficient:
(2x)ln(4) = (x-1)ln(9)
3. Expand the equation by distributing the logarithms:
2x*ln(4) = x*ln(9) - ln(9)
4. Move all terms containing x to one side of the equation:
2x*ln(4) - x*ln(9) = -ln(9)
5. Factor out x from the left side of the equation:
x(2ln(4) - ln(9)) = -ln(9)
6. Divide both sides of the equation by (2ln(4) - ln(9)) to solve for x:
x = -ln(9) / (2ln(4) - ln(9))
After obtaining the value of x using logarithms, you can plug it back into the original equation 4^(2x) = 9^(x-1) to check if it satisfies the equation.