Solving Exponential Functions - College Algebra

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Question

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

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Answer

Case 1,

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

This is true for all real values of , therefore,

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by , you would obtain $\ln(a)=\ln(b)$ which would imply which is not true for this case. We are solving for $a\neq b$ .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as $a \neq b$ .

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Question

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Tap to reveal answer

Answer

Case 1,

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

This is true for all real values of , therefore,

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by , you would obtain $\ln(a)=\ln(b)$ which would imply which is not true for this case. We are solving for $a\neq b$ .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as $a \neq b$ .

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Question

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Tap to reveal answer

Answer

Case 1,

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

This is true for all real values of , therefore,

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by , you would obtain $\ln(a)=\ln(b)$ which would imply which is not true for this case. We are solving for $a\neq b$ .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as $a \neq b$ .

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Question

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Tap to reveal answer

Answer

Case 1,

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

This is true for all real values of , therefore,

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by , you would obtain $\ln(a)=\ln(b)$ which would imply which is not true for this case. We are solving for $a\neq b$ .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as $a \neq b$ .

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Question

Find the value of a:

$\log_3(a) = 5$

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Answer

For this problem, we can use the fact that the exponent and logarithm are inverses of each other. So if we raise both sides up as exponents:

$3^{\log_3(a)} = 3^5$

The left side of the equation cancels and we get that

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Question

$5^{x}=2e^5$

Solve for .

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Answer

The first thing we notice about this problem is that is an exponent. This should be an immediate reminder: use logs!

The question is, which base should we choose for the log? We should use the natural log (log base e) because the right-hand side of the equation already has e as a base of an exponent. As you will see, things cancel out more nicely this way.

$5^{x}=2e^5$

Take the natural log of both sides:

$\textup{ln}(5^x) = \textup{ln}(2e^5)$

Rewrite the right-hand side of the equation using the product rule for logs:

$\textup{ln}5^x = \textup{ln}2+\textup{ln}e^5$

Now rewrite the whole equation after bringing down those exponents.

$x\textup{ln}5 = \textup{ln}2+5\textup{ln}e$

$\textup{ln}e$ is the same thing as $\textup{log}_{e}e$ , which equals 1.

$x\textup{ln}5 = \textup{ln}2+5$

Now we just divide by $\textup{ln}5$ on both sides to isolate .

$x = \frac{\textup{ln}2+5}{\textup{ln}5}$

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Question

Solve $6e^{-x}+1=3$ . Round to the nearest thousandth.

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Answer

The original equation is:

$6e^{-x}+1=3$

Subtract from both sides:

$6e^{-x}=2$

Divde both sides by :

$e^{-x}=\frac{2}{6}=\frac{1}{3}$

Take the natural logarithm of both sides:

$-x=\ln \frac{1}{3}$

Divde both sides by and use a calculator to get:

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Question

Solve for :

.

If necessary, round to the nearest tenth.

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Answer

Give both sides the same base, using e:

$e^{ln(x+3)}=e^3$ .

Because e and ln cancel each other out, .

Solve for x and round to the nearest tenth:

$x\approx17.1$

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Question

Solve the following for x:

$e^{x+2}=3$

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Answer

To solve this exponential function, you must first "undo" the e by taking the natural log of both sides. Thus,

$\ln(e^{x+2})=\ln3$

$x+2=\ln3$

Then, to isolate x, you must subtract two from both sides. Thus,

$x=\ln3-2$

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Question

Solve the following for x:

$2e^{4x}=4$

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Answer

To solve, you must isolate x. The first two steps are to divide both sides by 2 and then take the natural log of both sides.

$e^{4x}=2$

$\ln{e^{4x}}=\ln2$

$4x=\ln2$

Finally, you must divide both sides by 4.

$\frac{4x}{4}=\frac{\ln2}{4}$

$x=\frac{\ln2}{4}$

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Question

Solve for , round to the nearest hundredth.

$7(5^{y+1})=7^y$

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Answer

Solve for ,

$7(5^{y+1})=7^y$

Note the rules for logarithms,

$\ln(ab) = \ln(a)+\ln(b)$ (1)

$\ln\left(\frac{a}{b} \right )= \ln(a)-\ln(b)$ (2)

$\ln(a^n)=n\ln(a)$ (3)

There is more than one way to get started, but notice that if we divide out by we can proceed as follows,

$5^{y+1}=7^{y-1}$

Take the natural log of both sides,

$\ln(5^{y+1})=\ln(7^{y-1})$

Use rule (3) to pull out the exponents,

$(y+1)\ln(5)=(y-1)\ln(7)$

Expand with the distributive property,

$y\ln(5)+\ln(5)= y\ln(7)-\ln(7)$

Collect terms with on onside and then factor out .

$y[\ln(5)-ln(7)]=-ln(5)-ln(7)$

Divide out to solve for ,

$y = \frac{-\ln(5)-\ln(7)}{\ln(5)-\ln(7)}$ (4)

The numerator can be simplified using rules (2) and (3). Use rule (3) on the $-\ln(5)$ term in the numerator as follows,

$-\ln(5)= (-1)\ln(5)=\ln(5^{-1})=\ln\left(\frac{1}{5} \right )$

So now the numerator can be written,

$\ln\left(\frac{1}{5} \right ) -\ln(7) = \ln\left(\frac{1/5}{7} \right ) = \ln\left(\frac{1}{35} \right )=-\ln(35)$

Bringing back the denominator in equation (4) gives,

$y=-\frac{ \ln(35) }{\ln(5)-\ln(7)}$

$y \approx 10.57$

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Question

Solve this equation: $2^{x+4}=512$

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Answer

To solve $2^{x+4}=512$ , use the one-to-one property of exponential equations.

$2^{x+4}=2^9$

$\boldsymbol{x=5}$

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Question

Solve for .

$4^{3x+1}=256$

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Answer

To more easily solve for x, we need to try to express the right hand side of the equation in terms of a number of base 4. 256 is simply $4^{4}$ .

Rewriting the equation with this discovery yields

$4^{3x+1}=4^{4}$

Since both bases are the same, we can set the exponents equal to each other and solve for x.

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Question

Solve for .

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Answer

To solve for , set the two inside quantities equal to each other and solve.

$10^{log(-3x-1)}=10^{log(-4x-6)}$

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Question

Solve for .

$4^{2x}=4^{-2x-1}$

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Answer

Since the bases of the exponents are the same (4), the exponents have to equal each other for the exponential equation to be true.

$x=-\frac{1}{4}$

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Question

Solve for x:

$2^{x}=8$

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Answer

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this, factor the larger number until it looks like the smaller one. For this problem that looks as follows:

$2^{x}=8$

Factor the larger number:

$8=2^{3}$

Now substitute in the factored number:

$2^{x}=2^{3}$

Since the bases are now the same:

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Question

Solve for x in the exponent:

$3^{x}=\frac{1}{3}$

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Answer

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this with a fraction, factor the denominator then use the laws of exponents to eliminate the fraction. For this problem that looks as follows:

$3^{x}=\frac{1}{3}$

Manipulate the fraction:

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

Now substitute in the factored number:

$3^{x}=3^{-1}$

Since the bases are now the same:

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Question

Solve for x:

$4^{x}=64$

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Answer

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this, factor the larger number until it looks like the smaller one. For this problem that looks as follows:

$4^{x}=64$

Factor the larger number:

$64=4\cdot16\rightarrow4\cdot 4^{2}=4^{3}$

Now substitute in the factored number:

$4^{x}=4^{3}$

Since the bases are now the same:

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Question

Solve for x:

$2^{x}=32$

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Answer

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this, factor the larger number until it looks like the smaller one. For this problem that looks as follows:

$2^{x}=32$

Factor the larger number:

$32=4\cdot 8=2^{2}\cdot 2^{3}=2^{5}$

Now substitute in the factored number:

$2^{x}=2^{5}$

Since the bases are now the same:

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Question

Solve for x:

$3^{x}=81$

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Answer

In order to solve for x when it is in the exponent, we need to make the bases on either side of the equation look the same. In order to do this, factor the larger number until it looks like the smaller one. For this problem that looks as follows:

$3^{x}=81$

Factor the larger number:

$81=9^{2}=(3^{2})^{2}=3^{4}$

Now substitute in the factored number:

$3^{x}=3^{4}$

Since the bases are now the same:

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