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3 years ago

3^x+2=8^x-1 Solve for x.

Mathematics

1 answer:

zubka84 [21]3 years ago

5 0

I assume the equation is supposed to be

$3^{x+2}=8^{x-1}$

Then we can write

$9\cdot3^x=\dfrac18\cdot8^x\implies\left(\dfrac38\right)^x=\dfrac1{72}$

Take the logarithm of base 3/8 on both sides:

$\log_{3/8}\left(\dfrac38\right)^x=\log_{3/8}\dfrac1{72}$

$\implies x\log_{3/8}\dfrac38=-\log_{3/8}72$

$\implies x=-\log_{3/8}72$

- - -

If the equation is actually $3^x+2=8^x-1$ , I'm afraid it cannot be solved exactly.

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3 years ago

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Josh's textbook reached the ground first

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Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

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