Question

How to solve 5.012x=10(2x-1)?

Answer 1

`x=0.667200427`

Explanation:

Using BIDMAS, expanding out the brackets is the first step.

#color(red)(5.012x)=color(green)10(color(blue)(2x)color(green )(-1))#

`color(green)(10) xx color(blue)(2x)=color(blue)(20x)`

`color(green)(10) xx color(green)(-1)=color(green)(-10)`

Collecting like terms:

`color(red)(5.012x)=color(blue)(20x)color(green)(-10)`

We need to isolate the `color(blue)(x)` to one side, so we take `color(blue)(20x)` from both sides.

`color(red)(5.012x)color(blue)(-20x)=color(red)(-14.988x`

This leaves us with:

`color(red)(-14.988x)=color(green)(-10)`

To solve an equation divide the value `color(green)((-10))` by how many `color(red)(x's` we have `color(red)((-14.988)`

`therefore` `color(red)(x)=color(green)(-10)/color(blue)(-14.988)`

`color(red)(x)=color(red)((0.667200427...)`

`therefore` `x=0.667200427`

Answer 2

`color(red)( x = (-1) / ( log_10 (5.012) - 2 ) approx 0.76924`

Explanation:

If you solving: `5.012^x = 10^(2x-1)`

Taking `log_10` on both side:

`=> log_10 5.012^x = log_10 10^(2x-1)`

Now we must use our log laws:

`logbeta^alpha = alpha log beta`

`=> x log_10 5.012 = (2x-1)log_10 10`

We know `log_10 10 = 1`

`=> x log_10 5.012 = 2x-1`

Rearranging...

`=> x ( log_10 (5.012) - 2 ) = -1`

`color(red)( x = (-1) / ( log_10 (5.012) - 2 ) approx 0.76924`