college algebra problem

[coder1]

For the life of me I can't remeber how to start an equation with X in the exponent on both sides

example: 5^x=4^(x+1)

Thanks for any help

 

[eakers]

xln(5) = (x+1)ln4

 

[coder1]

Thats what I come up with too, but how do I get the (X+1) on the other side of the equation

 

[Packin]

Take the ln() of both sides, and then use the logarithm rule which states ln(x^y) = yln(x).

Edit: Wow, I was way too slow.

 

[Howard]

Originally posted by: coder1
Thats what I come up with too, but how do I get the (X+1) on the other side of the equation

Take the ln of both sides and use the power law.

 

[Heisenberg]

Originally posted by: coder1
Thats what I come up with too, but how do I get the (X+1) on the other side of the equation

Divide.

 

[coder1]

Ok so this is what I come up with when I divide.

(x ln 5)/(x+1)=ln 4

Is this right?

 

[Maximus96]

ln(1)=x/(x+1) , whoops, that should be ln(1)=(x+1)/x

 

[Maximus96]

ln != log

 

[Martin]

Originally posted by: coder1
Thats what I come up with too, but how do I get the (X+1) on the other side of the equation

Originally posted by: eakers
xln(5) = (x+1)ln4

xln5 = xln4 + ln4
x(ln5-ln4) = ln4
x = ln4/(ln5-ln4)
x = 6.212......

 

[AkumaX]

yeah, ln(x) = log(base e)(x) and log(x) = log(base 10)(x), also lg(x) = log(base 2)(x)

 

[coder1]

Thanks everyone, it was the factoring of x that threw me off

 

[EmperorIQ]

Originally posted by: AkumaX
yeah, ln(x) = log(base e)(x) and log(x) = log(base 10)(x), also lg(x) = log(base 2)(x)

^wow, did not know that

 

[Cogman]

However, in this problem, ln or log will work as the basic rule states log(base a) b^c = c*log(base a)( b ), ln is easier to write, and usually supported on most calculators.