# How to solve this log equation?

• homeworkhelpls
In summary: Change of base formula)In summary, the conversation discusses using the laws of logarithms to simplify an equation involving logarithmic expressions. The first step is to transform the equation using the laws of logarithms, and then simplify it using algebraic techniques. The final goal is to get an equation in the form of ##\log_a x = \log_a y##, which can be solved by taking the base to the power of both sides. The conversation also mentions the change of base formula, which can be used to convert logarithms to different bases.
homeworkhelpls

I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.

homeworkhelpls said:
View attachment 322211
I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.
If you have transformed ##\log(6x)-\log(18)## to ##\log(6x/18)## then why did you stop? Put in ##x-1## as well.

Btw.: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

fresh_42 said:
If you have transformed ##\log(6x)-\log(18)## to ##\log(6x/18)## then why did you stop? Put in ##x-1## as well.

Btw.: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
i mean i did transform the equation but after idk how to go on

homeworkhelpls said:
i mean i did transform the equation but after idk how to go on
Merge ##\log\left(\dfrac{6x}{18}\right)+\log(x-1)##. Then you get an equation ##\log \ldots = \log \ldots## which you can take ##b## to the power of it.

All terms are to the same base b. Properly using the logarithm properties and some simplifications should bring you to a step showing 3(x+4)=2x(x-1) .

mcastillo356
Laws of Logarithms
If ##x>0##, ##y>0##, ##a>0##, ##b>0##, ##a\neq 1##, and ##b\neq 1##, then
(i) ##\log_a 1=0##
(iii)##\log_a {(xy)}=\log_a x+\log_a y##
(iii)##\log_a {\left(\dfrac{1}{x}\right)}=-\log_a x##
(iv)##\log_a {\left(\dfrac{x}{y}\right)}=\log_a x-\log_a y##
(v)##\log_a {(x^y)}=y\log_a x##
(vi)##\log_a x=\displaystyle\frac{\log_b x}{\log_b a}##

nuuskur, MatinSAR, neilparker62 and 2 others

## 1. How do you solve a log equation?

To solve a log equation, you need to use the properties of logarithms to isolate the variable and solve for its value. This typically involves using the power rule, product rule, or quotient rule.

## 2. What are the properties of logarithms?

The properties of logarithms include the power rule, product rule, quotient rule, and change of base rule. These rules allow us to manipulate logarithmic expressions and solve equations involving logarithms.

## 3. Can you give an example of solving a log equation?

Sure! Let's say we have the equation log(x) + log(2) = log(8). We can use the product rule to combine the two logarithms on the left side, giving us log(2x) = log(8). Then, using the power rule, we can rewrite this as 2x = 8. Finally, we solve for x by dividing both sides by 2, giving us x = 4.

## 4. What if the log equation has a variable in the exponent?

If the log equation has a variable in the exponent, we can use the power rule to bring the exponent down and rewrite the equation without the exponent. Then, we can solve using the same steps as before.

## 5. Are there any common mistakes to avoid when solving log equations?

Yes, there are a few common mistakes to avoid when solving log equations. These include forgetting to apply the properties of logarithms, not checking for extraneous solutions, and mistaking logarithmic expressions for exponential expressions.

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