Solve for x in equation with subtraction of logs

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Clodagh Carroll

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Clodagh Carroll 1 year, 2 months ago

Published this.

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Clodagh Carroll 1 year, 2 months ago

Created this as a copy of Equation with subtraction of logs - with progressive hints.

Name	Status	Author	Last Modified
Solve for x in equation with addition of logs	draft	Clodagh Carroll	30/01/2024 12:33
Solve for x in equation with subtraction of logs	draft	Clodagh Carroll	30/01/2024 12:34

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Question statement

Solve for $x$ in the following equation, rounding your answer to $\var{decimal_places}$ decimal place(s):

$\log_{\var{base}}(\var{a}x+\var{b}) - \log_{\var{base}}(\simplify{{c}x+{n*d}}) = \var{f}$

Give any introductory information the student needs.

			Name	Type	Generated Value

base	integer	5
a	integer	2
b	integer	6
c	integer	7
d	integer	4
f	integer	2
g	number	25
n	integer	1
decimal_places	integer	2
x	number	-0.5433526012
numerator	number	94
denominator	number	-173
ans	number	-0.54

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Base of log terms in the equation.

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Generated value: `integer`

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Beginning with the equation given:

$\log_{\var{base}}(\var{a}x+\var{b}) - \log_{\var{base}}(\simplify{{c}x+{n*d}}) = \var{f} ,$

note the following rule of logs that allows us to convert between the log of a quotient and the subtraction of two separate log terms:

$\begin{align*} \log_a \left( \frac{x}{y} \right) & = \log_a x - \log_a y\\ \\ \text{e.g., } \log_3 \left( \frac{x}{9} \right) & = \log_3 (x) - \log_3 (9) \end{align*}$

Remember, we can go from left to right or from right to left with this rule. The important thing is that the $\textbf{BASE}$ is the same in the two log terms we are subtracting.

Applying this to the left-hand side of the equation above:

$\begin{align*} \underbrace{\color{black}{\log_{\var{base}}(\var{a}x+\var{b}) - \log_{\var{base}}(\simplify{{c}x+{n*d}})}} &= \var{f}\\ \searrow \\ & \log_{\var{base}}(\var{a}x+\var{b}) - \log_{\var{base}}(\simplify{{c}x+{n*d}}) =\log_{\var{base}} \left(\frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}}\right) \end{align*}$

In other words,

$\log_{\var{base}}\left(\frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}}\right)= \var{f}$

Next, note the following rule of logs that allows us to cancel a base and a log to the same base:

$\begin{align*} a^{\log_a (x)} & = x\\ \\ \text{e.g., } 3^{\log_3 (7)} & = 7 \end{align*}$

Therefore, to cancel out the $\textbf{LOG}$ to base $\var{base}$ , we must take both sides as a $\textbf{POWER}$ on $\var{base}$ :

$\begin{align*}\log_{\var{base}} \left( \frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}}\right) & = \var{f}\\ \phantom{.}^{\var{base}^{(\ \ \ )}} \bigg\downarrow & \hspace{0.5cm} \bigg\downarrow \phantom{.}^{\var{base}^{(\ \ \ )}} \\ \var{base}^{\log_{\var{base}} \left( \frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}} \right)} & = \var{base}^{\var{f}} \end{align*}$

We can now use rule of logs mentioned above to cancel the $\log_{\var{base}}$ with the base $\var{base}$ on the left-hand side:

$\begin{align*} \var{base}^{\log_{\var{base}} \left( \frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}}\right)} & = \var{base}^{\var{f}} \\ \frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}} & = \var{base}^{\var{f}} \end{align*}$

In other words,

$\frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}} = \simplify{{base}^{f}}$

We can now get rid of the fraction by $\textbf{MULTIPLYING}$ both sides by the denominator $\simplify{{c}x+{n*d}}$ :

$\begin{align*} \frac{\var{a}x+\var{b}}{\simplify{{c}x+{n*d}}} \cdot (\simplify{{c}x+{n*d}}) & = \simplify{{base}^{f}} \cdot (\simplify{{c}x+{n*d}})\\ \var{a}x+\var{b} & = \simplify{{base}^{f}} \cdot (\simplify{{c}x+{n*d}}) \end{align*}$

Now let's $\textbf{MULTIPLY OUT}$ the brackets on the right-hand side:

$\var{a}x+\var{b} = \simplify{{base}^{f}*{c}*x+{base}^{f}*{n*d}}$

PLAN:

Group all $x$ -terms on one side

and all number terms on the other side.

BE VERY CAREFUL WITH THE SIGNS HERE!

$\begin{align*} \simplify[!noLeadingMinus,unitFactor]{{a}*x-{{base}^{f}*{c}}*x} &= \simplify[!noLeadingMinus]{{{base}^{f}*{n*d}}-{b}}\\ \simplify[unitFactor]{{a-base^f*c}*x} & = \simplify{{base^f*n*d-b}}\\ \\ \text{Now, if there is a number in front of } & x, \text{ we can divide both sides by this number to get: }\\ \\ x & = \frac{\var{numerator}}{\var{denominator}}\\ & \hspace{-1cm}\phantom{.}^{\text{Calculator work}} \bigg\downarrow\\ x &= \var{ans} \end{align*}$

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Solve for x in equation with subtraction of logs

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Clodagh Carroll 1 year, 2 months ago

Clodagh Carroll 1 year, 2 months ago

Error in variable testing condition

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Solve for x in equation with subtraction of logs

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Clodagh Carroll 1 year, 2 months ago

Clodagh Carroll 1 year, 2 months ago

Error in variable testing condition

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Warning

Generated value: integer

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Objectives

Penalties

Generated value: `integer`