How do you solve 7^x=4^(x+2)7x=4x+2?

1 Answer
Mar 16, 2016

You must convert to logarithmic form.

Explanation:

7^x = 4^(x + 2)7x=4x+2

log7^x = log4^(x + 2)log7x=log4x+2

You must now use the rule loga^x = xlogalogax=xloga to simplify further.

xlog7 = (x + 2)log4xlog7=(x+2)log4

xlog7 = xlog4 + 2log4xlog7=xlog4+2log4

xlog7 - xlog4 = log4^2xlog7xlog4=log42

x(log7 - log4) = log16x(log7log4)=log16

You can now simplify even further by using the rule log_an - log_am = log_a(n / m)loganlogam=loga(nm)

x = log16/(log(7/4))x=log16log(74)

This is as simplified as it gets. Make sure to ask your teacher if they prefer your answer to be in logarithmic form (as shown above) or as an approximate value (rounded to an x number of decimal places).

Hopefully this helps!