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

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  • Thread starter tmt1
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In summary, Tim is asking for help in solving the equation [a^(3x+1)][b^(2x-2)]=c where a and b are positive real numbers and c is a positive constant. The conversation includes suggestions to take the logarithm of both sides and to use the properties of logarithms to simplify the equation. Tim also shares a potential solution, but is unsure how to proceed since there are two different numbers in the brackets.
  • #1
tmt1
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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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  • #2
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...
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
\(\displaystyle (3x+1)\ln a + (2x-2)\ln b=\ln c \)

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

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

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

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

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

What is an equation with powers?

An equation with powers is a mathematical expression that contains variables raised to different powers. These types of equations can be solved by using algebraic techniques such as factoring, the quadratic formula, or the laws of exponents.

How do I solve an equation with powers?

To solve an equation with powers, you need to isolate the variable on one side of the equation and simplify the other side. This can be done by using inverse operations, such as taking the square root to cancel out an exponent, or by factoring out common terms.

Can an equation with powers have more than one solution?

Yes, an equation with powers can have multiple solutions. This means that there can be more than one value for the variable that satisfies the equation. For example, a quadratic equation can have two solutions and a cubic equation can have three solutions.

Are there any special rules for solving equations with powers?

Yes, there are some special rules that can help simplify the process of solving equations with powers. These include the power rule, which states that when multiplying two terms with the same base, you can add the exponents, and the zero power rule, which states that any number raised to the power of zero is equal to one.

What are some common mistakes to avoid when solving equations with powers?

Some common mistakes to avoid when solving equations with powers include forgetting to apply the distributive property, not using the correct order of operations, and incorrectly applying the laws of exponents. It is important to double-check your work and make sure all steps are completed accurately.

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