How do you solve #13^(x-3) = 7^(-8x)#?

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Answer 1 · 2016-03-05T13:33:42

x = #(3log13)/ (log13+8log7)# = 0.424374, nearly.

Equate logarithms ( of any base ) and solve the resulting linear equation in x.

Answer 2 · 2016-03-05T14:08:47

#color(red)(" A very detailed explanation!")#

#" "x=(3log(13))/(8log(91))" "->" "color(blue)(x=(3color(white)(.)root(91)(13))/8)#

#color(green)(~~0.2132" to 4 decimal places")#

#color(blue)("Principle used to solve this problem")#

If you have #log(a^b)# then this can be written as #blog(a)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:#" "13^(x-3)=7^(-8x)#

#color(brown)("Take logs of both sides:")#

#" "log(13^(x-3))" " =" "log (7^(-8x)) #

#" "=> (x-3)log(13)" "=" "-8xlog(7)#

#color(brown)("Divide both sides by "log(13))#

#" "x-3" "=" "-8x xx(log(7))/(log(13)#

#color(brown)("Divide both sides by "-8x)#

#" "x/(-8x) -3/(-8x)" "=" "(log(7))/(log(13)#

#" "-1/8 +3/(8x)" "=" "(log(7))/(log(13)#

#color(brown)("Add "1/8" to both sides")#

#" "3/(8x)" "=" "(log(7))/(log(13)) + 1/8#

#color(brown)("Divide both sides by 3")#

#" "1/(8x)" " =" " (log(7))/(3log(13)) + 1/24#

#color(brown)("Multiply both sides by 8")#

#" "1/x" "=" "(8log(7))/(3log(13)) + 8/24#

#color(brown)("But "8/24 = 1/3 )#

#" " 1/x = (8log(7))/(3log(13)) + 1/3#

#" " 1/x=(8log(7)+log(13))/(3log(13))" "->"Corrected at this point"#

#color(brown)("Inverting everything")#

#" "x= (3log(13))/(8log(7)+log(13)#

#color(white)(.)#

#color(Red)("Corrected value for "x)#
#x~~ 3.3418/7.8747 = 0.4244# to 4 decimal places