How do you write #64=4^x# in Logarithm form?

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Answer 1 · 2016-09-01T12:54:14

#log_4 64 = x#

(this follows from the basic definition of log)

Answer 2 · 2016-09-01T13:04:55

#log_color(red)(a) color(blue)(b) = color(lime)(c) hArr color(red)(a)^color(lime)(c) = color(blue)(b)#

#log_color(red)(4) color(blue)(64) = color(lime)(x) hArr color(red)(4)^color(lime)(x) = color(blue)(64)#

"The #color(red)("base")# stays the #color(red)("base")#, and the other two change around"

log form and index form are interchangeable:

By definition: #log_color(red)(a) color(blue)(b) = color(lime)(c) hArr color(red)(a)^color(lime)(c) = color(blue)(b)#

Remember the following:

"The #color(red)("base")# stays the #color(red)("base")#, and the other two change around"

#log_color(red)(4) color(blue)(64) = color(lime)(x) hArr color(red)(4)^color(lime)(x) = color(blue)(64)#

#log_color(red)(4) color(blue)(64) = color(lime)(3) hArr color(red)(4)^color(lime)(3) = color(blue)(64)#