I have an equation and I've simplified it to this:
$$\frac {24}{19}=\frac {1.01583^{12x}}{1.01583^{12x}-1}$$
I'm having trouble finding $x$? I realize I have to find the log at some point, but am struggling as to when and where I do so?
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1$\begingroup$ can you use $\LaTeX$ please? $\endgroup$– Dr. Sonnhard GraubnerAug 31, 2016 at 14:09
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5$\begingroup$ Let $z=1.01583^{12x}$. Solve for $z$. Then take logs. $\endgroup$– luluAug 31, 2016 at 14:10
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2 Answers
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let $$a=1.01583^{12x}$$ then we have to solve $$\frac{24}{19}=\frac{a}{a-1}$$ from here we get $$24a-24=19a$$ thus we have $$a=\frac{24}{5}$$ can you proceed? from here we get $$\ln(a)=12x\ln(1.01583)$$ thus $$x=\frac{\ln(a)}{12\ln(1.01583)}$$
answered Aug 31, 2016 at 14:14
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Generalize the problem, solve $x$ when all the variables are real and positive:
$$\text{a}=\frac{\text{b}^{\text{c}x}}{\text{b}^{\text{c}x}-\text{d}}\Longleftrightarrow\text{a}=\frac{\frac{\text{b}^{\text{c}x}}{\text{b}^{\text{c}x}}}{\frac{\text{b}^{\text{c}x}}{\text{b}^{\text{c}x}}-\frac{\text{d}}{\text{b}^{\text{c}x}}}\Longleftrightarrow$$ $$\text{a}=\frac{1}{1-\frac{\text{d}}{\text{b}^{\text{c}x}}}\Longleftrightarrow\frac{1}{\text{a}}=1-\frac{\text{d}}{\text{b}^{\text{c}x}}\Longleftrightarrow$$ $$\frac{1}{\text{a}}-1=-\frac{\text{d}}{\text{b}^{\text{c}x}}\Longleftrightarrow1-\frac{1}{\text{a}}=\frac{\text{d}}{\text{b}^{\text{c}x}}\Longleftrightarrow$$ $$\frac{\text{d}}{1-\frac{1}{\text{a}}}=\text{b}^{\text{c}x}\Longleftrightarrow\frac{\log_{\text{b}}\left(\frac{\text{d}}{1-\frac{1}{\text{a}}}\right)}{\text{c}}=x$$
So, for your prolem we get:
$$x=\frac{\log_{1.01583}\left(\frac{1}{1-\frac{1}{\frac{24}{19}}}\right)}{12}=\frac{\log_{1.01583}\left(\frac{24}{5}\right)}{12}\approx8.322799606$$
answered Aug 31, 2016 at 14:20