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

\(\displaystyle 4=2^2, 8=2^3\) therefore

\(\displaystyle (2^2)^{2x+5}=(2^3)^{x}\) Apply power rule of exponents.

\(\displaystyle 2^{4x+10}=2^{3x}\) With the same base, we can now write

\(\displaystyle 4x+10=3x\) Subtract \(\displaystyle 4x\) on both sides.

\(\displaystyle 10=-x\) Divide \(\displaystyle -1\) on both sides.

\(\displaystyle -10=x\)

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

\(\displaystyle 25=5^2, 125=5^3\) therefore

\(\displaystyle (5^2)^{x+1}=(5^3)^{x+4}\) Apply power rule of exponents.

\(\displaystyle 5^{2x+2}=5^{3x+12}\) With the same base, we can now write

\(\displaystyle 2x+2=3x+12\) Subtract \(\displaystyle 2x, 12\) on both sides.

\(\displaystyle x=-10\)

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:  \(\displaystyle \frac{4}{37}\)