# -s6.6 solve exponential eq 3^x-14\cdot 3^{-x}=5

• MHB
• karush
In summary, the conversation discusses solving the exponential equation $3^x-14\cdot 3^{-x}=5$. One method is to rewrite the equation as $3^x+\dfrac{14}{3^x}=5$ and then solve the resulting quadratic equation $3^{2x}-5\cdot3^x-14 =0$. Another method is to let $y=3^x$ and solve the resulting quadratic equation $y^2-5y-14=0$. The only solution for $3^x$ is 7, giving a final solution of $x=\log_3(7)=\dfrac{\log 7}{\log 3}\approx 1.7712 karush Gold Member MHB \tiny{s464\\6.6} solve the exponential equation.$3^x-14\cdot 3^{-x}=5$Rewrite$\quad 3^x+\dfrac{14}{3^x}=5\times\quad 3^x \quad 3^{2x}+14=5\cdot 3^x$quadratic$\quad 3^{2x}-5\cdot3^x-14 =0$Factor$\quad (3^x-7)(3^x+2)=0$Discard$\quad 3^x =7$hence$\quad x=\log_3(7)=\dfrac{\log 7}{\log 3}\approx 1.7712$ok i think it is correct probably better to use () rather than$\cdot\$
didn't know is factoring out the 3 was possible

That's good. I would have done it just slightly differently. I would have let y= 3^x so that the equation became y- 14/y= 5. multiplying by y gives y^2- 5y- 14= 0. That, of course, factors as (y- 7)(y+ 2)= 0 so that y= 7 or y= -2. But 3^x cannot be negative so the only solution is 3^x= 7, x log(3)= log(7) and x= log(7)/log(3) as you had.

(I don't see where you "factored out" a 3.)

## What is an exponential equation?

An exponential equation is an equation in which the variable appears in the exponent. It can be written in the form of y = ab^x, where a and b are constants and x is the variable.

## How do I solve an exponential equation?

To solve an exponential equation, you can use logarithms or rewrite the equation in a form that is easier to solve. In this case, we can rewrite the equation as 3^x = 5 + 14*3^(-x) and then use substitution to solve for x.

## What is the solution to the equation 3^x-14*3^(-x)=5?

The solution to this equation is x = 1. This can be found by rewriting the equation as 3^x = 5 + 14*3^(-x) and then using substitution to solve for x.

## What is the relationship between exponential equations and logarithms?

Logarithms are the inverse operation of exponential equations. This means that if you have an equation in the form of y = ab^x, you can rewrite it as x = log_b(y/a). This relationship can be used to solve exponential equations.

## Are there any practical applications for solving exponential equations?

Yes, exponential equations are commonly used in fields such as finance, biology, and physics. They can be used to model population growth, compound interest, and radioactive decay, among other things.

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