(Exponential Equation) How do I find X?

#5^x=4^(x+1)#

2 Answers
Noah G
Apr 13, 2017

#x~~ 6.21#.

Explanation:

Take the natural logarithm of both sides.

#ln(5^x) = ln(4^(x +1))#

Now use #lna^n = nlna#.

#xln5 = (x + 1)ln4#

#xln5 = xln4 + ln4#

#xln5 - xln4 = ln4#

#x(ln5 - ln4) = ln4#

This can be simplified further using #lna -lnb = ln(a/b)#.

#x(ln(5/4)) = ln4#

#x = (ln4)/(ln(5/4))#

If you prefer an approximation, we can take #x~~ 6.21#.

Hopefully this helps!

Gió
Apr 13, 2017

I got: #x=(ln(4))/(ln(5)-ln(4))#

Explanation:

Here we can try applying the natural log, #ln#, on both sides and apply some properties of logs:
#ln(5)^x=ln(4)^(x+1)#
then:
#xln(5)=(x+1)ln(4)#
rearrange:
#xln(5)=xln(4)+ln(4)#
#xln(5)-xln(4)=ln(4)#
#x[ln(5)-ln(4)]=ln(4)#
#x=(ln(4))/(ln(5)-ln(4))#
if you have a pocket calculator we can easily evaluate the natural log to get:
#x=6.21256#

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