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ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1360188674673:dw

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1360189814146:dw

robtobey
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{The} \text{ derivative} \text{ of } 5^{x } \text { is}: \]\[5^x\text{ Log}[5] \]

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0not sure i can do this because (5^x) (ln5) separate

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0then integral of u is (5^x)/ln5 right

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1360189966266:dw

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0if i use u substitution ln5 goes to top so no use

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large \int 5^x\ln 5 dx \] ?? yes `

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large \ln 5 \int 5^xdx = \ln 5 \int e^{\ln5x}dx \]

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0If I understand the integral correctly.

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0Just rewrite the number 5 as e^ln5 and then you have a regular exponential function (linear) you can substitute and evaluate.

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0can u really do that manuplation

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0Of course, here's the proof: \[\Large a^b=c \\ \\ \Large b\log_aa= \log_ac\]that leads to : \[\Large b=\log_ac \] And back substitution gives: \[\Large a^{\log_ac}=c \] Just a different way of writing numbers, that's also the reason why a lot of people write the exponential decay with as an e function, rather then the regular y(t)=ba^t form

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.0Integration result gives you \[\Large e^{\ln5 x}+C \] Deriving with respect to x: \[\Large \ln5 \cdot e^{\ln 5 x} = \ln 5 \cdot \left(e^{\ln5} \right)^x = \ln5 \cdot 5^x\]
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