Solve using logarithm:
2^(4x+1)=3^x
take logs of both sides
log 2^(4x+1) = log 3^x
use rules of logs
(4x+1)log2 = xlog3
4xlog2 + log2 = xlog3
4xlog2 - xlog3 = -log2
x(4log2 - log3) = -log2
x = -log2/(4log2 - log3)
x = log2/(log3 - 4log2)
= ..
let the button-pushing begin
Thank you!
To solve the equation 2^(4x+1) = 3^x using logarithms, we need to use the property of logarithms that says:
If a^b = c, then log_a(c) = b.
Applying this property, we can take the logarithm of both sides of the equation to get:
log_2(2^(4x+1)) = log_2(3^x)
Using the power rule of logarithms, the left side becomes:
(4x+1) * log_2(2) = x * log_2(3)
Since log_2(2) = 1, the equation simplifies to:
4x + 1 = x * log_2(3)
Now, we can isolate the variables by subtracting 4x from both sides:
1 = x * log_2(3) - 4x
Moving the terms involving x to one side gives:
x * log_2(3) - 4x = 1
Factor out x:
x * (log_2(3) - 4) = 1
Finally, divide both sides by (log_2(3) - 4) to solve for x:
x = 1 / (log_2(3) - 4)
This is the exact solution to the equation using logarithms. If you want a decimal approximation, you can substitute the value of log_2(3) (approx. 1.58) into the equation.