Solve for x in equation with addition of logs

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Clodagh Carroll

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Clodagh Carroll 1 year, 2 months ago

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Clodagh Carroll 1 year, 3 months ago

Created this as a copy of Solve for x in equation with addition of logs.

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:

$\log_{\var{base}}(x+\var{a}) + \log_{\var{base}}(\simplify{x+{n*b}}) = 2 \log_{\var{base}}(\simplify{x+{m*c}})$

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			Name	Type	Generated Value

base	integer	9
a	integer	2
b	integer	8
c	integer	4
n	integer	-1
m	integer	-1
x	number	16
numerator	number	32
denominator	integer	2

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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:

$\log_{\var{base}}(x+\var{a}) + \log_{\var{base}}(\simplify{x+{n*b}}) = 2 \log_{\var{base}}(\simplify{x+{m*c}}) ,$

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

$\begin{align*} \log_a (xy) & = \log_a x + \log_a y\\ \\ \text{e.g., } \log_2 (7x) & = \log_2 (7) + \log_2 (x) \end{align*}$

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 adding.

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

$\begin{align*} \underbrace{\log_{\var{base}}(x+\var{a}) + \log_{\var{base}}(\simplify{x+{n*b}})} &= 2 \log_{\var{base}}(\simplify{x+{m*c}})\\ \searrow \\ & \log_{\var{base}}(x+\var{a}) + \log_{\var{base}}(\simplify{x+{n*b}}) =\log_{\var{base}} [(x+\var{a})(\simplify{x+{n*b}})] \end{align*}$

In other words,

$\log_{\var{base}}[(x+\var{a})(\simplify{x+{n*b}})] = 2 \log_{\var{base}}(\simplify{x+{m*c}})$

Now, for the right-hand side, note the following rule of logs:

$\hspace{0.5cm} \begin{align*} \log_a (x^p) &= p \cdot \log _a (x)\\ \\ \text{e.g., } \log_3 (7^4) & = 4 \cdot \log_3(7) \end{align*} \hspace{0.5cm}$

Applying this to the right-hand side of our equation:

$\begin{align*} \log_{\var{base}}[(x+\var{a}) (\simplify{x+{n*b}})] = \underbrace{2 \log_{\var{base}}(\simplify{x+{m*c}})} &\\ \searrow \\ & 2 \log_{\var{base}}(\simplify{x+{m*c}}) =\log_{\var{base}}[(\simplify{x+{m*c}})^2] \end{align*}$

In other words,

$\log_{\var{base}}[(x+\var{a}) (\simplify{x+{n*b}})] = \log_{\var{base}}[(\simplify{x+{m*c}})^2]$

We can now $\textbf{"CANCEL"}$ the $\log_{\var{base}}$ on both sides and this gives:

$(x+\var{a}) (\simplify{x+{n*b}}) = (\simplify{x+{m*c}})^2$

Next, $\textbf{MULTIPLY OUT}$ the brackets on both sides:

$\begin{align*} \underbrace{(x+ \var{a}) (\simplify{x+{n*b}})} & = \underbrace{(\simplify{x+{m*c}})^2}\\ \hspace{-2cm} \bigg\downarrow & \phantom{\hspace{1.5cm}} \searrow\\ x(\simplify{x + {n*b}}) +\var{a}(\simplify{x+{n*b}})& = \underbrace{(\simplify{x+{m*c}})(\simplify{x+{m*c}})} \\ \hspace{-2.5cm} \bigg\downarrow & \phantom{\hspace{1.8cm}}\bigg\downarrow\\ \simplify{x^2 + {n*b}*x} +\simplify{{a}x+{a*n*b}} & =\simplify[!noLeadingMinus]{x^2+{m*c}*x+{m*c}*x+{m*c}^2}\\ \simplify{x^2+{a+n*b}*x+{a*n*b}} & = \simplify{x^2+{2*m*c}*x+{(m*c)^2}} \end{align*}$

Next, we can subtract $x^2$ from both sides:

$\begin{align*} \simplify{x^2+{a+n*b}*x+{a*n*b}} -x^2 & = \simplify{x^2+{2*m*c}*x+{(m*c)^2}} -x^2\\ \simplify{{a+n*b}*x+{a*n*b}} & = \simplify[!noLeadingMinus]{{2*m*c}*x+{(m*c)^2}} \end{align*}$

PLAN:

Group all $x$ -terms on one side

and all number terms on the other side.

BE VERY CAREFUL WITH THE SIGNS HERE!

$\begin{align*} \simplify[!noLeadingMinus,unitFactor]{{a+n*b}*x-{2*m*c}*x} &= \simplify[!noLeadingMinus]{{(m*c)^2}-{a*n*b}}\\ \simplify[unitFactor]{{a+n*b-2*m*c}*x} & = \simplify{{(m*c)^2-{a*n*b}}}\\ \\ \text{Now, if there is a number in front of } & x, \text{ we can divide both sides by this number to get: }\\ \\ x & = \simplify{{numerator}/{denominator}}\end{align*}$

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Solve for x in equation with addition of logs

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Clodagh Carroll 1 year, 2 months ago

Clodagh Carroll 1 year, 3 months ago

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Solve for x in equation with addition 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, 3 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`