Question

Hi i need help with 15a(ii)

Answer 1

5/2

Explanation:

So first of all 1/log_3(4) can be written as log_4(3)...

So everything is base 4 except the log_8(x^3)...

We can play with this to get it so that the base is 4.

Let y=log_8(x^3) then 8^y=x^3

Rewrite 8 as 4^(3/2) so we have

4^(3/2 *y)=x^3

Now rewriting in log form gives: log_4(x^3)=3/2*y

Then solving that for y gives 2/3*log_4(x^3) or log_4(x^2)... let's put it back into the equation:

log_4(x^2+5x)-log_4(x^2)=log_4(3)

log_4((x^2+5x)/x^2)=log_4(3)

Set insides equal:

(x^2+5x)/x^2=3

Cross multiply:

x^2+5x=3x^2

Subtract 3x^2 on both sides:

-2x^2+5x=0

Factor

-x(2x-5)=0

So solutions are 0 and 5/2.

We have to verify these...

0 isn't going to work because we can't do log of 0

it makes x^2+5x 0 and x^3 0

The only solution is 5/2.

Answer 2

x = 5/2

Explanation:

log4(x^2+5x)-log8(x^3)=1/log3(4)

log(x^2 + 5 x) / log(4) - log(x^3) / log(8) = log(3) / log(4)

log(x (x+5))/log(4) - log(x^3) / log(8) = log(3) / log(4)

(3 log(x (x+5)) - 2 log(x^3)) / log(64) = log(3) / log(4)

3 log(x (x+5)) - 2 log(x^3) = 3 log(3)

log((3 x)/(x+5))=0

x=5/2