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As explained on the Converting between Exponential and Logarithmic Form page, logarithmic expressions can be expressed in two equivalent forms - *exponential form* and *logarithmic form*. We can use this to our advantage when solving logarithmic/exponential equations, especially when asked to find exact value solutions (as we'll demonstrate in the follow examples).

Recall that, by definition, we have:

\(y=b^x \iff x=log_b(y)\)

Example: Solve each of the following exponential equations by converting to logarithmic form, leaving answers as exact values.

a) \(3^x=7\) |
b) \(6^{3y}=4\) |
c) \(e^{2k-1}=5\) |

**Solutions:**

**a) **Converting \(3^x=7\) into logarithmic form gives \(x=log_3(7)\) as the solution.

**b)** Converting \(6^{3y}=4\) into logarthmic form gives \(3y=log_6(4)\). Then, we can isolate for \(y\) by dividing both sides by \(3\), so we have \(y=\frac{log_6(4)}{3}\) as the solution.

**c) **Converting \(e^{2k-1}=5\) into logarithmic form gives \(2k-1=log_e(5)=ln(5)\). Then, we can isolate for \(k\) by adding \(1\) to both sides, then dividing both sides by \(2\), so that we get \(k=\frac{ln(5)+1}{2}\) as the solution.

Example: Solve each of the following logarithmic equations by converting to exponential form, leaving answers as exact values.

a) \(log_7(343)=x\) |
b) \(5log_2(y)=40\) |
c) \(log_2(log_2(w))=3\) |

**Solutions:**

**a)** Converting \(log_7(343)=x\) into exponential form gives \(7^x=343\). Since \(7^3=343\), we have \(7^x=7^3\), so equating exponents gives us the solution \(x=3\).

**b) **We first divide both sides of \(5log_2(y)=40\) to find the logarithmic expression that we can convert to exponential form, so we have:

\(log_2(y)=\frac{40}{5}=8\)

Then, converting to exponential form and solving for \(y\) gives us:

\(y=2^8 \Rightarrow \) \(y=256\)

**c) **We have nested logarithms on the left-hand side, so we will need to apply the exponential form conversion twice to find \(w\):

\(log_2(log_2(w))=3\)

\(\Rightarrow log_2(w)=2^3=8\)

\(\Rightarrow w=2^8\)

\(\Rightarrow\) \(w=256\)

Another method we can use to solve exponential equations is to take the logarithm of both sides of the equation, then use the properties/rules or identities of logarithms to solve them.

*__Note__: We can take the logarithm in any base and still end up with the __same__ solution/result (demonstrated in the following example).

Example: Solve the equation \(3^{x-2}=8\) as an exact value and an approximate value rounded to 2 decimal places, by first taking the logarithm of both sides of the equation in:

a) base \(10\) |
b) base \(e\) |
c) base \(3\) |

**Solution:**

**a) base \(10\) (common log)**

Taking the logarithm in base \(10\) on both sides of the equation gives:

\(log(3^{x-2})=log(8)\)

Then after applying the logarithm power rule to the left side, we have:

\((x-2)log(3)=log(8)\)

Finally, dividing both sides by \(log(3)\) then adding \(2\) to both sides gives us the solution:

\(x-2=\frac{log(8)}{log(3)}\)

\(\Rightarrow\) \(x=\frac{log(8)}{log(3)}+2 \approx 3.89\)

**b) base \(e\) (natural log or \(ln\))**

Following the same steps as in part **a) **but with base \(e\), we have the following calculations:

\(ln(3^{x-2}\)=ln(8)\)

\(\Rightarrow (x-2)ln(3)=ln(8)\)

\(\Rightarrow x-2=\frac{ln(8)}{ln(3)}\)

So, we have the solution:

\(x=\frac{ln(8)}{ln(3)}+2 \approx 3.89\)

**c) base \(3\)**

Taking the logarithm in base \(3\) on both sides of the equation gives:

\(log_3(3^{x-2})=log_3(8)\)

Since the logarithm base is the same as the base of the exponent, we can use the logarithm property/identity \(log_b(b^x)=x\) to simplify the left-hand side of the equation so we have:

\(x-2=log_3(8)\)

So the solution to the equation is:

\(x=log_3(8)+2 \approx 3.89\)

**Tip:** Taking the logarithm of both sides in the same base as the base of the exponent (like we did in part **c)** of the previous example) allows us to use the logarithm property/identity \(log_b(b^x)=x\) to further simplify our solutions.

We can also use the properties/rules and identities from the Applying Logarithm Rules to Simplify Expressions page to help us simplify equations to make them easier to solve.

Example: Solve the equation \(2log(3)+\frac{1}{2}log(16)-log(3)=log(x)\).

**Solution:**

We solve this equation by simplifying it using the following logarithm properties/rules and identities:

\(2log(3)+\frac{1}{2}log(16)-log(3)=log(x)\)

\(\Rightarrow log(3^2)+log(16^{\frac{1}{2}})-log(3)=log(x)\) Power Rule

\(\Rightarrow log(9)+log(4)-log(3)=log(x)\) Evaluating powers in the logarithms

\(\Rightarrow log(9\times 4)-log(3)=log(x)\) Product Rule

\(\Rightarrow log(36)-log(3)=log(x)\) Evaluating power in the logarithm

\(\Rightarrow log(\frac{36}{3})=log(x)\) Quotient Rule

\(\Rightarrow log(12)=log(x)\) Evaluating power in the logarithm

Finally, equating the powers in the logarithms on both sides of the equation gives us the solution:

\(x=12\)

When solving logarithmic equations, it is important to check that the solutions we find are not extraneous ones. Recall that we can only take the logarithm of positive values, but it is actually possible to find solutions that, when substituted back into the original equation, are not in the domain of the logarithmic function. We will see this occur with the following example:

Example: Solve the equation \(log_3(3x)+log_3(x-2)=2\).

**Solution:**

Using the Product Rule, we can simplify the expression on the left side of the equation into one logarithm:

\(log_3(3x(x-2))=2\)

Converting this equation into exponential form gives:

\(3x(x-2)=3^2=9\)

Then, expanding and rearranging this equation results in the quadratic equation:

\(3x^2-6x-9=0\)

Factoring the left-side and solving gives:

\(3(x^2-2x-3)=3(x-3)(x+1)=0\)

\(\Rightarrow x=3, x=-11\)

Although these values are solutions to the quadratic equation, they are not necessarily solutions to the original equation \(log_3(3x)+log_3(x-2)=2\). So now, we check by substituting them into the left side of the original equation to see if it's equal to the right side:

\(x=3\) | \(x=-1\) |

\(log_3(3(3))+log_3((3)-2)\) \(=log_3(9)+log_3(1)\) \(=2+0\) \(=2\) \(\Rightarrow\) \(x=3\) is a solution |
\(log_3(3(-1))+log_3((-1)-2)\) \(=log_3(-3)+log_3(-3)\) \(\leftarrow\) the value of \(log_3(-3)\) is undefined since \(-3\) is not in the domain of the function \(f(x)=log_3(x)\)
\(\Rightarrow\) \(x=-1\) is |

So, the *only* solution of the original equation \(log_3(3x)+log_3(x-2)=2\) is \(x=3\).

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Aug 2, 2022 11:43 AM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=718098
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