What is x

An MaxWong,

thanks for the hint.
I did not want to have a term with "lg" in the result.

In addition, I was "persecuted" and had to get ready quickly.

Greeting asinus :- ) !

7log(3x)+3log(5x)=15 what is x

i assume log is the natural logarithm

$\begin{array}{|rcll|} \hline 7\cdot log(3x)+3 \cdot log(5x) &=& 15 \\ 7\cdot (~log(3)+log(x) ~)+3 \cdot (~log(5)+log(x) ~) &=& 15 \\ 7\cdot log(3) + 7\cdot log(x) +3\cdot log(5) + 3\cdot log(x)&=& 15 \\ 10\cdot log(x) + 7\cdot log(3) +3\cdot log(5) &=& 15 \\ 10\cdot log(x) + log(3^7) + log(5^3) &=& 15 \\ 10\cdot log(x) + log(3^7\cdot 5^3) &=& 15 \quad & | \quad -log(3^7\cdot 5^3) \\ 10\cdot log(x) &=& 15 -log(3^7\cdot 5^3) \quad & | \quad : 10 \\ log(x) &=& \frac{ 15 -log(3^7\cdot 5^3) } {10 } \\ log(x) &=& \frac{ 15}{10} - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \log(~(3^7\cdot 5^3)^{\frac{1}{10}}~) \\ log(x) &=& 1.5 - \log(3^\frac{7}{10}\cdot 5^\frac{3}{10}) \\ log(x) &=& 1.5 - \log(3^{0.7}\cdot 5^{0.3}) \quad & | \quad e^{()} \\ x &=& e^{1.5 - \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot e^ {- \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot \frac{ 1 } { e^{\log(3^{0.7}\cdot 5^{0.3}) } } \\ x &=& e^{1.5} \cdot \frac{ 1 } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ e^{1.5} } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ 4.48168907034 } {2.15766927997\cdot 1.62065659669 } \\ x &=& \frac{ 4.48168907034 } {3.49684095207 } \\ \mathbf{x} & \mathbf{=} & \mathbf{1.28163938016178...} \\ \hline \end{array}$

7log(3x)+3log(5x)=15 what is x

i assume lg is the 10 - logarithm

$7(lg3+lgx)+3(lg5+lgx)=15$

$7lg3+7lgx+3lg5+3lgx=15$

$7lgx+3lgx=15-(7lg3+3lg5)$

$lgx^7+lgx^3=15-lg \ 273375$

$lg(x^7\times x^3)=15-lg \ 273375$

$lgx^{10}=15-lg \ 273375$

$lgx^{10}+lg273375=15$

$lg \ (273375 \ x^{10})=15$

$273375 x^{10}=10^{15}$

$\large x^{10}=\frac{10^{15}}{273375}$

$\Large x=\sqrt[10]{\frac{10^{15}}{273375}}$

$\large x=9.04324132413$

  !

7log(3x)+3log(5x)=15 what is x

$7(lg3+lgx)+3(lg5+lgx)=15$

$\large x=9.04324132413$

$\large 10.03412+4.96588=15$

  !