Vereinfache folgende Ausdrücke
(b,c,xb,c,xb,c,x und yyy seien positiv mit b,c≠1b,c\ne 1b,c=1)
bx+logbyb^{x+\log_by}bx+logby
Für diese Aufgabe benötigst Du folgendes Grundwissen: Rechnen mit Logarithmus
bx+logby= bx ⋅ blogby=bx ⋅ ylogbb=bx ⋅ yb^{x+\log_by}=\ b^x\ \cdot\ b^{\log_by}=b^x\ \cdot\ y^{\log_bb}=b^x\ \cdot\ ybx+logby= bx ⋅ blogby=bx ⋅ ylogbb=bx ⋅ y
Kommentiere hier 👇
alogcb=alogablogac=(alogab)1logac=b1logac=blogcalogcc=blogcaa^{\log_cb}=a^{\frac{\log_ab}{\log_ac}}=(a^{\log_ab})^{\frac{1}{\log_ac}}=b^{\frac{1}{\log_ac}}=b^{\frac{\log_ca}{\log_cc}}=b^{\log_ca}alogcb=alogaclogab=(alogab)logac1=blogac1=blogcclogca=blogca
(b)logbx\left(\sqrt[]{b}\right)^{\log_bx}(b)logbx
((b)logbx = xlogbb = xlogb(b12) = x12⋅logbb = x\left((\sqrt[]{b}\right)^{\log_bx}\ =\ x^{\log_b\sqrt[]{b}}\ =\ x^{\log_b\left(b^{\frac{1}{2}}\right)}\ =\ x^{\frac{1}{2}\cdot\log_bb}\ =\ \sqrt[]{x}((b)logbx = xlogbb = xlogb(b21) = x21⋅logbb = x
logc(x1logcb)\log_c\left(x^{\frac{1}{\log_cb}}\right)logc(xlogcb1)
logc(x1logcb) =1logcb ⋅ logcx = logbx \log_c\left(x^{\frac{1}{\log_cb}}\right)\ =\frac{1}{\log_cb}\ \cdot\ \log_cx\ =\ \log_bx\ logc(xlogcb1) =logcb1 ⋅ logcx = logbx
Bitte melde dich an, um diese Funktion zu benutzen.