Wie löse ich die folgenden Potenzgleichungen?
3∙5^2x=7^(x+4)
Logarithmieren:
log10(3)*2x*log10(5)=(x+4)*log10(7)
Und dann? Wie stelle ich es um, damit ich x errechnen kann?
5∙8^(x+1)=〖16〗^(x-1)
Logarithmieren:
log10(5)*(x+1)*log10(8)=(x-1)log10(16)
Und dann= Wie stelle ich es um, damit ich x errechnen kann?
+26387 5∙8^(x+1)=〖16〗^(x-1)
$$\begin{array}{rcl}
5\cdot 8^{(x+1)} &=& 16^{(x-1)} \\
5\cdot 8^x\cdot 8^1 &=& 16^x \cdot 16^{-1} \\
40\cdot 8^x &=& \dfrac{16^x}{16}\\\\
40\cdot 16 &=& \dfrac{16^x}{8^x}\\\\
\dfrac{16^x}{8^x} &=& 40\cdot 16 \\\\
\dfrac{16^x}{8^x} &=& 640\\\\
\left( \dfrac{16}{8} \right)^x &=& 640\\\\
2^x &=& 640 \quad | \quad \log_{10}\\\\
\log_{10}(2^x) &=& \log_{10}(640) \quad | \quad \log_{10}\\\\
x\cdot \log_{10}(2) &=& \log_{10}(640) \\\\
x &=& \dfrac{ \log_{10}(640) } { \log_{10}(2) }\\\\
x &=& \dfrac{ 2.80617997398} {0.30102999566}\\\\
x &=& 9.32192809489
\end{array}$$
+26387 Wie löse ich die folgenden Potenzgleichungen?
3∙5^2x=7^(x+4)
$$\begin{array}{rcl}
3\cdot5^{2x} &=& 7^{(x+4)}\\
3\cdot 5^{2x} &=& 7^x\cdot 7^4\\
\dfrac{5^{2x} }{7^x} &=& \dfrac{7^4}{3} \quad | \quad \log_{10}\\\\
\log_{10}\left(
\dfrac{5^{2x} }{7^x} \right)
&=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
\log_{10}\left( 5^{2x} \right) - \log_{10}\left( 7^x \right) &=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
2x\cdot \log_{10}\left( 5\right) - x \cdot \log_{10}\left( 7\right)
&=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
x \cdot \left[ 2\cdot \log_{10}( 5 )- \log_{10}( 7 ) \right]
&=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
x \cdot \left[ \log_{10}( 5^2 )- \log_{10}( 7 ) \right]
&=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
x \cdot \left[ \log_{10} \left( \dfrac{5^2}{ 7 }\right) \right]
&=&
\log_{10}\left(
\dfrac{7^4}{3} \right) \\ \\
x &=& \dfrac{ \log_{10} \left( \dfrac{7^4}{ 3 }\right) }
{ \log_{10}\left( \dfrac{5^2}{7} \right) } \\ \\
x &=& \dfrac{ \log_{10} \left( 800.333333333 \right) }
{ \log_{10}\left( 3.57142857143 \right) } \\ \\
x &=& \dfrac{ 2.90327090534 }
{ 0.55284196866 } \\ \\
x &=& 5.25153854073
\end{array}$$
+26387 5∙8^(x+1)=〖16〗^(x-1)
$$\begin{array}{rcl}
5\cdot 8^{(x+1)} &=& 16^{(x-1)} \\
5\cdot 8^x\cdot 8^1 &=& 16^x \cdot 16^{-1} \\
40\cdot 8^x &=& \dfrac{16^x}{16}\\\\
40\cdot 16 &=& \dfrac{16^x}{8^x}\\\\
\dfrac{16^x}{8^x} &=& 40\cdot 16 \\\\
\dfrac{16^x}{8^x} &=& 640\\\\
\left( \dfrac{16}{8} \right)^x &=& 640\\\\
2^x &=& 640 \quad | \quad \log_{10}\\\\
\log_{10}(2^x) &=& \log_{10}(640) \quad | \quad \log_{10}\\\\
x\cdot \log_{10}(2) &=& \log_{10}(640) \\\\
x &=& \dfrac{ \log_{10}(640) } { \log_{10}(2) }\\\\
x &=& \dfrac{ 2.80617997398} {0.30102999566}\\\\
x &=& 9.32192809489
\end{array}$$
+14538 $${\mathtt{x}} = {\frac{{log}_{10}\left({\mathtt{640}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}}$$ => $${\frac{{log}_{10}\left({\mathtt{640}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}} = {\mathtt{9.321\: \!928\: \!094\: \!887\: \!361\: \!6}}$$
$${\frac{{log}_{10}\left({\frac{{{\mathtt{7}}}^{{\mathtt{4}}}}{{\mathtt{3}}}}\right)}{{log}_{10}\left({\frac{{{\mathtt{5}}}^{{\mathtt{2}}}}{{\mathtt{7}}}}\right)}} = {\mathtt{5.251\: \!538\: \!540\: \!726\: \!350\: \!5}}$$
!
