Question 268217
How do you solve the exponential equation in this problem?
3^x=243


If we take the log (base 10) of both sides we have:

log (base 10) 3^x = log (base 10) 243

Using the law that says log a^b = b*log a we have:

x*log (base 10) 3 = log (base 10) 243

x = [log (base 10) 243]/[log (base 10) 3]

x = [log (base 10 (2.43*100)]/[log (base 10) 3]

Using the law that says log (a*b) = log a + log b:

x = [log (base 10) 2.43 + log (base 10) 100]/[log (base 10) 3]

Since the log (base 10) of 100 = 2 (remember 10^2 = 100 so the log (base 10) of 100 = 2):

x = [(log (base 10) 2.43) + 2]/[log (base 10) 3.

Look up the base 10 logs for 2.43 and 3 and finish the calculation.