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

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To solve the equation [a^(3x+1)][b^(2x-2)] = c, taking the logarithm of both sides is recommended. This leads to the equation (3x+1)ln a + (2x-2)ln b = ln c. Assuming a and b are positive real numbers, the equation cannot equal zero. The final form allows for isolating x, but the presence of different logarithmic terms complicates the solution. The discussion emphasizes the importance of logarithmic properties in solving equations involving exponents.
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Hi,

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

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

Tim
 
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tmt said:
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...
 
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
 
Re: SImple question

tmt said:
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.
 
Re: SImple question

I like Serena said:
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
 
$$(3x+1)\ln a + (2x-2)\ln b=\ln c $$

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

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

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

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