
Haz la sustitución:


Para confirmar el resultado:


Sustituye:


(Te dejaré confirmar por ti mismo.)

Sustituye:



Sustituye:


Podemos hacer que esto se vea un poco mejor:


Take the logarithm of both sides. The base of the logarithm doesn't matter.


Drop the exponents:

Expand the right side:

Move the terms containing <em>x</em> to the left side and factor out <em>x</em> :


Solve for <em>x</em> by dividing boths ides by 5 log(4) - log(3) :

You can stop there, or continue simplifying the solution by using properties of logarithms:



You can condense the solution further using the change-of-base identity,

Answer:
- Gabe earns 13.5 per hour
x ------------------ 4 ---- 8 --------- 12 ------ 16
Earnings ----- 57 ---- 111 ------- 165------ 219
Step-by-step explanation:
Answer:
x=6
Step-by-step explanation:
To find the perimeter of the triangle you need to add up all the sides, so if all the sides are 2x-3 the equation will look like this;
2x-3+2x-3+2x-3=27
which when combining like terms will leave it to be;
6x-9=27
Than move all the factors besides x over to the right side;
6x-9+9=27+9
6x = 36
6x/6 = 36/6
x = 6
so your answer will be x=6