Question

How to use logarithm to solve for x?

Can someone please explain to me how to do from question e onwards? Thanks!

Answer 1

Each of these log questions are written in the form `a^(bx)=c`, as `log_a(b)^c-=clog_a(b)` (the power rule), then `log(a)^(bx)=log(c)-=bxlog(a)=log(c)`, where `x=log(c)/(blog(a))`," (change of base rule).

For example, let's take `2^(5x)=100`. We can take the log of both sides to get `log(2)^(5x)=log(100)`, using the power law we get: `5xlog(2)=log(100)`, so `x=log(100)/(5log(2))=1.32877124~~1.33`

Answer 2

`(e)color(white)(x) 1+log_(10)11`

Explanation:

`(e)`

`"using the "color(blue)"laws of logarithms"`

`•color(white)(x)logx^nhArrnlogx`

`•color(white)(x)log_b b=1`

`"given "10^x=110`

`"take " log_(10)" of both sides"`

`rArrlog_(10)10^x=log_(10)110`

`rArrxcancel(log_(10)10)^1=log_(10)(11xx10)`

`rArrx=log_(10)10+log_(10)11`

`rArrx=1+log_(10)11`