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Olenka [21]
3 years ago
12

Explain how solve 4^(x+3)=7 using the change of base formula log base b of y equals log y over log b. Include the solution for x

in your answer. Round your answer to the nearest thousandth.
Mathematics
1 answer:
Naily [24]3 years ago
8 0

The value of x is -1.596

<em><u>Solution:</u></em>

<em><u>Given equation is:</u></em>

4^{(x+3)} = 7

Let us solve using change of base formula log base b of y equals log y over log b

From given,

4^{(x+3)} = 7

\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)

Therefore,

\ln \left(4^{x+3}\right)=\ln \left(7\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\ln \left(4^{x+3}\right)=\left(x+3\right)\ln \left(4\right)\\\\\left(x+3\right)\ln \left(4\right)=\ln \left(7\right)\\

Let us simplify the above

\left(x+3\right)\cdot \:2\ln \left(2\right)=\ln \left(7\right)\\\\\mathrm{Divide\:both\:sides\:by\:}2\ln \left(2\right)\\\\\frac{\left(x+3\right)\cdot \:2\ln \left(2\right)}{2\ln \left(2\right)}=\frac{\ln \left(7\right)}{2\ln \left(2\right)}\\\\

\mathrm{Simplify}\\\\x+3=\frac{\ln \left(7\right)}{2\ln \left(2\right)}\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x+3-3=\frac{\ln \left(7\right)}{2\ln \left(2\right)}-3\\\\\mathrm{Simplify}\\\\x=\frac{\ln \left(7\right)}{2\ln \left(2\right)}-3

Substitute the values

ln 7 = 1.9459101490553132

ln 2 = 0.6931471805599453

Therefore,

x = \frac{1.9459101490553132}{2 \times 0.6931471805599453} - 3\\\\x = 1.40367746103 - 3\\\\x = -1.59632253897 \approx -1.596

Thus solution for x is found

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<h3>Answer: point Q is located at (-1, 1)</h3>

=================================================

Explanation:

Check out the diagram below.

Plot R(-3,7) and T(3,-11) on the same xy grid.

Draw a vertical line through R and a horizontal line through T. A right triangle forms. At the intersection point is point S(-3,-11)

----------------------

Now measure the distance from point S to point T. You can count out the spaces or subtract the x coordinates and use absolute value.

|x1-x2| = |-3-3| = |-6| = 6

From point S to point T is 6 units.

We want to subdivide this horizontal length in the ratio 1:2

What this means is that we want to plot a point U somewhere such that SU:UT = 1:2

In other words,

SU = x

UT = 2x

SU+UT = ST

x+2x = 6

3x = 6

x = 6/3

x = 2

So we must move 2 spaces to the right from point S to get to U(-1,-11)

Going from point U(-1,-11) to T(3,-11) is 4 spaces

We have SU:UT = 2:4 = 1:2 to help confirm we have the correct location for point U

From point U, we then move straight up to the line segment RT

We'll land on Q(-1,1)

----------------------

Another way to find the y coordinate of point Q is to subdivide the segment RS into the ratio 1:2 similar to how we divided ST up

Segment RS is 18 units long since we go from y = 7 to y = -11 when going from R to S.

If V was on segment RS such that

RV:VS = 1:2

and RV = y

then

RV+VS = RS

y+2y = 18

3y = 18

y = 6

RV = y = 6

VS = 2y = 2*6 = 12

So you'll move 6 units down from y = 7 to land on y = 1 (when going from the y coordinate of R to the y coordinate of Q)

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