When dealing with exponential equations, we want to make sure the bases are the same. This way we can set-up an equation with the exponents. Since the bases are now different, we need to convert so we have the same base. We do know that

$\displaystyle \frac{1}{2}=2^{-1}$ therefore

$\displaystyle 2^{x-9}=(2^{-1})^{x+1}$ With the same base and also applying power rule of exponents, we now can write

$\displaystyle x-9=-x-1$ Add $\displaystyle x, 9$ on both sides.

$\displaystyle 2x=8$ Divide $\displaystyle 2$ on both sides.

$\displaystyle x=4$

When dealing with exponential equations, we want to make sure the bases are the same. This way we can set-up an equation with the exponents. Since the bases are now different, we need to convert so we have the same base. We do know that

$\displaystyle \frac{1}{4}=4^{-1}$ therefore

$\displaystyle 4^{2x+3}=(4^{-1})^{-x+6}$ With the same base and by also applying the power rule of exponents, we can now write

$\displaystyle 2x+3=x-6$ Subtract $\displaystyle x, 3$ on both sides.

$\displaystyle x=-9$

In order to solve this equation, we will need to change the bases so that both bases are equal.  We can change the bases to base three.

$\displaystyle \frac{1}{3}=3^{-1}$

$\displaystyle \frac{1}{81} = 3^{-4}$

Rewrite the terms in the equation.

$\displaystyle (3^{-1})^{8x} = (3^{-4})^{3-x}$

The powers can be set equal to each other now that we have similar bases.

$\displaystyle (-1)(8x) = (-4)(3-x)$

Solve for x.

$\displaystyle -8x = -12+4x$

Subtract $\displaystyle 4x$ on both sides.

$\displaystyle -8x-4x = -12+4x-4x$

$\displaystyle -12x=-12$

Divide by negative 12 on both sides.

$\displaystyle \frac{-12x}{-12}=\frac{-12}{-12}$

The answer is:  $\displaystyle 1$

Evaluate by changing the base of the fraction.

$\displaystyle \frac{1}{16} = 16^{-1}$

Rewrite the equation.

$\displaystyle (16)^{-1(9x)} = 16^{3+x}$

With common bases, we can set the powers equal to each other.

$\displaystyle -1(9x)=3+x$

Solve for x.

$\displaystyle -9x = 3+x$

Subtract x from both sides.

$\displaystyle -9x -x= 3+x-x$

$\displaystyle -10x = 3$

Divide by negative ten on both sides.

$\displaystyle \frac{-10x }{-10}= \frac{3}{-10}$

The answer is:  $\displaystyle -\frac{3}{10}$

Convert the right side to a base of 7.

$\displaystyle 7^{x-1} = 7^{2(\frac{1}{x})}$

With similar bases, the powers can be set equal to each other.

$\displaystyle x-1 = \frac{2}{x}$

Multiply both sides by $\displaystyle x$.

$\displaystyle x(x-1 )=x( \frac{2}{x})$

$\displaystyle x^2-x =2$

Subtract two from both sides.

$\displaystyle x^2-x -2=2-2$

$\displaystyle x^2-x -2=0$

Factorize the left term to determine the roots of the quadratic.

$\displaystyle (x+1)(x-2)=0$

Set each binomial equal to zero and solve for x.

$\displaystyle x+1 =0, x=-1$

$\displaystyle x-2=0, x=2$

Ensure that both solutions will satisfy the original equation.  Since there is no zero denominator resulting from substitution, both answers will satisfy.

The answers are:  $\displaystyle -1,2$

Convert the right fraction to a base of two thirds.

$\displaystyle \frac{4}{9} =(\frac{2}{3})^2$

Rewrite the right side.

$\displaystyle (\frac{2}{3})^{-5x} =(\frac{2}{3})^{2(2+x)}$

With similar bases, set both powers equal to each other.

$\displaystyle -5x = 2(2+x)$

Simplify the right side by distribution.

$\displaystyle -5x= 4+2x$

Subtract $\displaystyle 2x$ from both sides.

$\displaystyle -5x-2x= 4+2x-2x$

$\displaystyle -7x =4$

Divide both sides by negative seven.

$\displaystyle \frac{-7x }{-7}=\frac{4}{-7}$

The answer is:  $\displaystyle -\frac{4}{7}$

We can rewrite the left side of the equation as follows:

$\displaystyle \frac{1}{e^{-3x+6}} =( \frac{1}{e})^{-3x+6} = (e^{-1})^{-3x+6}$

Rewrite the equation.

$\displaystyle (e^{-1})^{-3x+6} = e^{6x}$

Since both sides share the same bases, we can set the powers equal to each other.

$\displaystyle (-1)(-3x+6) =6x$

Solve for x.

$\displaystyle 3x-6 = 6x$

Subtract $\displaystyle 3x$ on both sides.

$\displaystyle 3x-6 -3x= 6x-3x$

$\displaystyle -6 =3x$

Divide by three on both sides.

$\displaystyle \frac{-6 }{3}=\frac{3x}{3}$

The answer is:  $\displaystyle -2$

Rewrite the right side using three as the base.

$\displaystyle (\frac{1}{81})^{9x-1} = (3^{-4})^{9x-1}$

The equation becomes:

$\displaystyle 3^x = (3)^{-4(9x-1)}$

Since both sides share similar bases, we can set the powers equal to each other.

$\displaystyle x=-36x+4$

Add $\displaystyle 36x$ on both sides.

$\displaystyle x+36x=-36x+4+36x$

$\displaystyle 37x = 4$

Divide both sides by 37.

$\displaystyle \frac{37x }{37}= \frac{4}{37}$

The answer is:  $\displaystyle \frac{4}{37}$

We can rewrite the right side by changing the base.

$\displaystyle \frac{1}{8} = (\frac{1}{2})^3$

Rewrite the right side using the new base.

$\displaystyle (\frac{1}{2})^{5x+2} =[(\frac{1}{2})^3]^{3x}$

Now that the bases are similar, we can set the exponents equal to each other.

$\displaystyle 5x+2 = 9x$

Subtract $\displaystyle 5x$ on both sides.

$\displaystyle 5x+2 -5x= 9x-5x$

$\displaystyle 2=4x$

Divide by four on both sides.

$\displaystyle \frac{2}{4}=\frac{4x}{4}$

Reduce the fractions.

The answer is:  $\displaystyle \frac{1}{2}$

We will need to convert the bases on both sides to base three.

$\displaystyle \frac{1}{3} = 3^{-1}$

$\displaystyle \frac{1}{9}=(\frac{1}{3})^2$

Rewrite the equation.

$\displaystyle 3[(3)^{-1}]^{3x+1}= (3^{-2})^{10}$

Simplify both sides.

$\displaystyle 3(3)^{-1(3x+1)}= (3^{-2})^{10}$

On the left side, notice that a lone three is the coefficient of the base.  We will need to write the left side such that:

$\displaystyle 3(3)^{-1(3x+1)} = 3^1\cdot (3)^{-1(3x+1)} = 3^{1+(-1)(3x+1)}$

Recall that if the powers of the same base are multiplied, the powers can be added.

Simplify the exponent.

$\displaystyle 3^{1+(-1)(3x+1)} = 3^{1-3x-1} = 3^{-3x}$

The equation becomes:  $\displaystyle 3^{-3x} = 3^{-20}$

Set the exponential terms equal to each other now that the bases are equal.

$\displaystyle -3x=-20$

Divide by negative three on both sides.

$\displaystyle \frac{-3x}{-3}=\frac{-20}{-3}$

The answer is:  $\displaystyle \frac{20}{3}$