MHB Solving Equations with Exponents: How to Find the Value of x

[tmt1]

Hi,

Is there a way to solve such an equation for x?

[a^(3x+1)][b^(2x-2)]Thanks,

Tim

 

[I like Serena]

Hi,

Is there a way to solve such an equation for x?

[a^(3x+1)][b^(2x-2)]Thanks,

Tim

Hi Tim!

What's the equation?
An equation is supposed to have a '=' sign in it...

 

[tmt1]

Re: SImple question

In this case, it could be any constant, either 0 or c, it is just because I am curious.

[a^(3x+1)][b^(2x-2)]=0

[a^(3x+1)][b^(2x-2)]=c

 

[I like Serena]

Re: SImple question

In this case, it could be any constant, either 0 or c, it is just because I am curious.

[a^(3x+1)][b^(2x-2)]=0

[a^(3x+1)][b^(2x-2)]=c

If we assume that $a$ and $b$ are positive real numbers, then it is not possible with 0.
That is because the power of a positive number is never 0.

With a positive constant $c$, I suggest you take the logarithm of both sides.
\begin{array}{}
a^{3x+1}b^{2x-2}&=& c \\
\ln \left( a^{3x+1}b^{2x-2}\right)&=&\ln c \\
(3x+1)\ln a + (2x-2)\ln b&=&\ln c
\end{array}
Perhaps you know how to solve that.

 

[tmt1]

Re: SImple question

If we assume that $a$ and $b$ are positive real numbers, then it is not possible with 0.
That is because the power of a positive number is never 0.

With a positive constant $c$, I suggest you take the logarithm of both sides.
\begin{array}{}
a^{3x+1}b^{2x-2}&=& c \\
\ln \left( a^{3x+1}b^{2x-2}\right)&=&\ln c \\
(3x+1)\ln a + (2x-2)\ln b&=&\ln c
\end{array}
Perhaps you know how to solve that.

Okay, thanks that is helpful but I do not know how to solve it.

I could solve

(3x+1)ln a + (3x+1) ln b = lnc

x=[ln c / (ln a + ln b) -1 ] / 3

However since there are 2 different number in the brackets I don't know what to do.

Thanks,

Tim

 

[Jameson]

$$(3x+1)\ln a + (2x-2)\ln b=\ln c $$

$$3x\ln a +\ln a+2x\ln b-2\ln b=\ln c$$

$$3x\ln a+2x\ln b=\ln c - \ln a+2\ln b$$

I think this should be a good hint but if not, see below. :)

Spoiler

$$x(3\ln a +2\ln b) = \ln c - \ln a+2\ln b $$