What value will x have to for the first number to reach zero 153•0.92^x

Step-by-step explanation:

153•0.92^x is a decaying exponential function.

In theory these functions never reach zero.

Suppose that you were willing to change the problem to read:

"What value will x have to for the first number to come within 0.0001 of zero?" Solve 153•0.92^x = 0.0001.

To do this, take the common log of both sides, obtaining

log 153 + x*log 0.92 = log 0.0001

Note that log 0.0001 = -4; log 153 = 2.18469; and log 0.92 = -0.03621.

Then we have:

2.18469 + x(-0.03621) = - 4.

Isolate the 2nd term.  To accomplish this, subtract 2.18469 from both sides, obtaining:

-0.03621x = -6.18469

Isolate x by dividing both sides by -0.03621:

x = 170.8

This tells us that as x approaches +179, the quantity 153·0.92x will be within 0.0001 of zero.

answer: i believe it’s 245.04

step-by-step explanation: