Transforming equation with $e$ in order to find $x$
$begingroup$
this task comes from electrotechnics.
How can I modify this equation in order to find $x$?
I need to have the equation in such form:
$x = text{...something...}$
Here is the equation:
$$expleft(-frac{a}{frac{1}{2}bx}right) = frac{2c-d}{d}$$
(this is e to the power of $-frac{a}{frac{1}{2}bx}$, looks confusing).
Thank you.
algebra-precalculus
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
$endgroup$
add a comment |
$begingroup$
this task comes from electrotechnics.
How can I modify this equation in order to find $x$?
I need to have the equation in such form:
$x = text{...something...}$
Here is the equation:
$$expleft(-frac{a}{frac{1}{2}bx}right) = frac{2c-d}{d}$$
(this is e to the power of $-frac{a}{frac{1}{2}bx}$, looks confusing).
Thank you.
algebra-precalculus
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
$endgroup$
3
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00
add a comment |
$begingroup$
this task comes from electrotechnics.
How can I modify this equation in order to find $x$?
I need to have the equation in such form:
$x = text{...something...}$
Here is the equation:
$$expleft(-frac{a}{frac{1}{2}bx}right) = frac{2c-d}{d}$$
(this is e to the power of $-frac{a}{frac{1}{2}bx}$, looks confusing).
Thank you.
algebra-precalculus
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
$endgroup$
this task comes from electrotechnics.
How can I modify this equation in order to find $x$?
I need to have the equation in such form:
$x = text{...something...}$
Here is the equation:
$$expleft(-frac{a}{frac{1}{2}bx}right) = frac{2c-d}{d}$$
(this is e to the power of $-frac{a}{frac{1}{2}bx}$, looks confusing).
Thank you.
algebra-precalculus
algebra-precalculus
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
edited Jan 20 at 0:45
David C. Ullrich
60.9k43994
60.9k43994
asked Jan 20 at 0:27
wenoweno
29211
asked Jan 20 at 0:27
wenoweno
29211
asked Jan 20 at 0:27
wenoweno
29211
29211
3
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00
add a comment |
3
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00
3
3
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Taking the natural logarithm of both sides gives:
$$frac{-2a}{bx} = ln(frac{2c-d}{d})$$
Then divide both sides by $ln(frac{2c-d}{d})$ and multiply both sides by $x$ to obtain:
$$x = frac{-2a}{bln(frac{2c-d}{d})}$$
answered Jan 20 at 1:48
user636441user636441
908
$endgroup$
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Taking the natural logarithm of both sides gives:
$$frac{-2a}{bx} = ln(frac{2c-d}{d})$$
Then divide both sides by $ln(frac{2c-d}{d})$ and multiply both sides by $x$ to obtain:
$$x = frac{-2a}{bln(frac{2c-d}{d})}$$
answered Jan 20 at 1:48
user636441user636441
908
$endgroup$
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
add a comment |
$begingroup$
Taking the natural logarithm of both sides gives:
$$frac{-2a}{bx} = ln(frac{2c-d}{d})$$
Then divide both sides by $ln(frac{2c-d}{d})$ and multiply both sides by $x$ to obtain:
$$x = frac{-2a}{bln(frac{2c-d}{d})}$$
answered Jan 20 at 1:48
user636441user636441
908
$endgroup$
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
add a comment |
$begingroup$
Taking the natural logarithm of both sides gives:
$$frac{-2a}{bx} = ln(frac{2c-d}{d})$$
Then divide both sides by $ln(frac{2c-d}{d})$ and multiply both sides by $x$ to obtain:
$$x = frac{-2a}{bln(frac{2c-d}{d})}$$
answered Jan 20 at 1:48
user636441user636441
908
$endgroup$
Taking the natural logarithm of both sides gives:
$$frac{-2a}{bx} = ln(frac{2c-d}{d})$$
Then divide both sides by $ln(frac{2c-d}{d})$ and multiply both sides by $x$ to obtain:
$$x = frac{-2a}{bln(frac{2c-d}{d})}$$
answered Jan 20 at 1:48
user636441user636441
908
edited Jan 20 at 20:00
answered Jan 20 at 1:48
user636441user636441
908
answered Jan 20 at 1:48
user636441user636441
908
answered Jan 20 at 1:48
user636441user636441
908
908
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
add a comment |
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
1
1
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so ln x gives $ln x$ compared to ln x which gives $ln x$
$endgroup$
– Ross Millikan
Jan 20 at 2:25
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
$begingroup$
Thanks Ross! Very helpful.
$endgroup$
– user636441
Jan 20 at 20:01
add a comment |
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3
$begingroup$
Take the natural log of both sides. Cross multiply and it's a linear equation in $x$.
$endgroup$
– Ethan Bolker
Jan 20 at 0:29
$begingroup$
What exactly is your difficulty with this problem? Is just that it "looks confusing"? This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
$endgroup$
– Rory Daulton
Jan 20 at 0:43
$begingroup$
Well what exactly have you tried?
$endgroup$
– Michael Wang
Jan 20 at 2:00