Question

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

Answer 1

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

Explanation:

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

Answer 2

`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")`

Explanation:

`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