What is the inverse of $y = 3 log (5 x) - ln (5 x^{3})$ $y=3log(5x)-ln(5x^3)$ ? ?

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What is the inverse of $y = 3 log (5 x) - ln (5 x^{3})$ $y=3log(5x)-ln(5x^3)$ ? ?

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Answer 1 · 2016-06-22T14:36:50

$y = 1.33274 \times 10^{\frac{- 0.767704 x}{3}}$ $y = 1.33274 xx10^((-0.767704 x)/3)$ for $0 < x < \infty$ $0 < x < oo$

Explanation:

Supposing that $log a = {log}_{10} a, ln a = {log}_{e} a$ $log a = log_{10}a, ln a = log_e a$

For $0 < x < \infty$ $0 < x < oo$

$y = \frac{{log}_{e} {(5 x)}^{3}}{{log}_{e} 10} - {log}_{e} {(5 x)}^{3} + {log}_{e} 25$ $y = log_e(5x)^3/log_e 10-log_e(5x)^3+log_e 25$

$y {log}_{e} 10 = (1 - {log}_{e} 10) {log}_{e} {(5 x)}^{3} + {log}_{e} 25 \times {log}_{e} 10$ $y log_e10 = (1-log_e10)log_e(5x)^3+log_e25 xxlog_e 10$

${log}_{e} {(5 x)}^{3} = \frac{y {log}_{e} 10 - {log}_{e} 25 \times {log}_{e} 10}{1 - {log}_{e} 10}$ $log_e(5x)^3=(y log_e10 - log_e25 xxlog_e 10)/ (1-log_e10)$

${(5 x)}^{3} = c_{0} e_{1}^{c_{1} y}$ $(5x)^3=c_0e^{c_1y}$

where $c_{0} = e^{- \frac{{log}_{e} 25 \times {log}_{e} 10}{1 - {log}_{e} 10}}$ $c_0 = e^(-(log_e25 xxlog_e 10)/ (1-log_e10))$
and $c_{1} = \frac{{log}_{e} 10}{1 - {log}_{e} 10}$ $c_1 = log_e10/(1-log_e10)$

Finally

$x = \frac{1}{5} c_{0}^{\frac{1}{3}} \times e^{\frac{c_{1}}{3} y}$ $x = 1/5 c_0^{1/3} xx e^{c_1/3 y}$

or

$x = 1.33274 \times 10^{\frac{- 0.767704 y}{3}}$ $x = 1.33274 xx10^((-0.767704 y)/3)$

Red $y = 3 log (5 x) - ln (5 x^{3})$ $y=3log(5x)-ln(5x^3)$
Blue $y = 1.33274 \times 10^{\frac{- 0.767704 x}{3}}$ $y = 1.33274 xx10^((-0.767704 x)/3)$

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What is the inverse of $y = 3 log (5 x) - ln (5 x^{3})$ $y=3log(5x)-ln(5x^3)$ ? ?

1 Answer

Explanation:

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What is the inverse of y=3log(5x)-ln(5x^3)y=3log(5x)−ln(5x3)y=3log(5x)-ln(5x^3)? ?

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What is the inverse of $y = 3 log (5 x) - ln (5 x^{3})$ $y=3log(5x)-ln(5x^3)$ ? ?