Moved - Logarithm Problem

[abel muroi]

what about 5^x = 4 ^{x + 1} ?

 

[pka]

what about 5^x = 4 ^{x + 1} ?

\(\displaystyle (\log(5))x=(2\log(2))(x+1)\)
Solve for \(\displaystyle x\).

 

[abel muroi]

\(\displaystyle (\log(5))x=(2\log(2))(x+1)\)
Solve for \(\displaystyle x\).

im stuck, not sure what to do next, can you give me a hint to point me in the right direction?

 

[abel muroi]

Abel, you're guessing. Get real:shock:
Substitute x=2 in that thing of yours:
5^2 = 4^(2+1)
5^2 = 4^3
25 = 64 : see what I mean?

Im not guessing, i got this problem from my textbook and im still stuck.

 

[daon2]

Im not guessing, i got this problem from my textbook and im still stuck.

When you're faced with \(\displaystyle a^{f(x)}=b^{g(x)}\) where \(\displaystyle a,b\) are positive real numbers, \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are polynomials, you want to turn an exponential equation into a polynomial equation by applying a logarithm. It becomes \(\displaystyle \log\left(a^{f(x)}\right)=\log\left(b^{g(x)}\right)\) or \(\displaystyle f(x)\cdot \log(a) = g(x)\cdot \log(b)\).

\(\displaystyle \log(a),\log(b)\) are just numbers, like \(\displaystyle -3,\pi,\sqrt{2}\). If you have trouble solving \(\displaystyle (x)\cdot \log(5)= (x+1)\cdot \log (4)\) then you're missing a pre-requisite for this and probably much of your future material. You might ask your teacher to go over it with you.

 

[abel muroi]

When you're faced with \(\displaystyle a^{f(x)}=b^{g(x)}\) where \(\displaystyle a,b\) are positive real numbers, \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are polynomials, you want to turn an exponential equation into a polynomial equation by applying a logarithm. It becomes \(\displaystyle \log\left(a^{f(x)}\right)=\log\left(b^{g(x)}\right)\) or \(\displaystyle f(x)\cdot \log(a) = g(x)\cdot \log(b)\).

\(\displaystyle \log(a),\log(b)\) are just numbers, like \(\displaystyle -3,\pi,\sqrt{2}\). If you have trouble solving \(\displaystyle (x)\cdot \log(5)= (x+1)\cdot \log (4)\) then you're missing a pre-requisite for this and probably much of your future material. You might ask your teacher to go over it with you.

its not that i am missing a pre requisite, its just that i don't remember completely, i took the "pre requisite" about two years ago so im a little rusty. can you give me a hint on how to solve that?

 

[daon2]

its not that i am missing a pre requisite, its just that i don't remember completely, i took the "pre requisite" about two years ago so im a little rusty. can you give me a hint on how to solve that?

Try to solve this for x: A*x=B*(x+1). Start by distributing the B on the right-hand side. Then move all x terms to one side, everything else to the other. Factor out the x, and divide both sides by everything that is beside x.

Now recognize that A can be replaced with log(5), B with log(4).

 

[stapel]

its not that i am missing a pre requisite, its just that i don't remember completely, i took the "pre requisite" about two years ago...

And, if you haven't used that material in the past two years, you've likely forgotten enough of it that you need to re-take those courses. Then you'll have the necessary background solidly in mind, allowing you to learn the new material.