7.07 Exponential and logarithmic equations

2^{1-2x}=\dfrac{1}{512}

The denominator of the right side of the equation, 512, is a power of 2. By first rewriting this as a power with a base of 2, we can simplify the process needed to solve this equation.

In general, a fraction in an exponential equation will typically lead to a negative exponent.

e^{2x+3}=4

We cannot rewrite 4 as e to some power, so we cannot use that method to solve this equation. The inverse of e^x is \ln\left(x\right), so we can use the equality property of logarithms to take the natural logarithm of both sides. Then, use the inverse property of logarithms to simplify.

The exact answer is x=\dfrac{\ln\left(4\right)-3}{2}.

We can find an approximate solution using technology, x\approx -0.8069.

2^{3x-1}=5^{x}

We cannot rewrite 2 and 5 as powers with the same base, and taking the logarithm of both sides would be complicated. Estimating the solution using technology is the simplest method for solving this equation.

First, we need to graph y=2^{3x-1} and y=5^x on the same plane.

Then, we need to adjust the window to view the intersection of the lines. The lines appear to intersect between x=1 and x=2 and between y=9 and y=12.

We can use the intersect feature of a graphing calculator to find that the lines intersect at x\approx 1.475.

Solving this equation by taking the logarithm of both sides is complicated but possible.

If we used a calculator to approximate the decimal, we would find x\approx1.475.

\dfrac{1}{3}\log_{5} x +7=8

The first thing we want to do is isolate the logarithm. Then, we can rewrite it in exponential form and solve for x. The domain of the logarithm is x>0.

The result lies within the domain, so the solution x=125 is valid.

When we get to the step \log_5 x =3, we can use the equality property of exponents instead of switching it to exponential form.

This method is helpful when solving logarithmic inequalities.

\log_6 \left(x\right)+\log_6 \left(x+9\right)=2

We can rewrite the left hand side using the product property of logarithms: \log_b\left(x\right)+ \log_b\left(y\right) =\log_b\left(xy\right) We can then use the identity \log_b x=n \iff x=b^n , to rewrite the expression without logarithms.

This gives us two possible solutions x=3 and x=-12.

To test for extraneous solutions, we want to check if any solution falls outside the domain of the logarithms. The logarithms in the original equation are \log_6 x and \log_6 \left(x+9\right), and we cannot take the logarithm of a negative value.

According to the first term, x must be greater than 0. According to the second term, x must be greater than -9. However, any number between -9 and 0 will make the first term undefined, so the domain for this equation is \left(0, \infty\right).

We now need to check if either solution is extraneous. As -12 < 0, it falls outside the domain of the function and is therefore extraneous.

x=3 is a valid solution.

We can check our solution by substituting x=3 into the equation:

Substituting x=3 makes the equation true, so we confirm that it is a valid solution.

\log _n\left(x+4\right)-\log _n\left(x-2\right)=\log _n\left(x\right)

We can rewrite the left hand side using the quotient property of logarithms: \log_b\left(x\right)- \log_b\left(y\right) =\log_b\left(\frac{x}{y}\right) This will give us an equation of the form \log_b\left(x\right)=\log_b\left(y\right) which we can simplify using the equality property\log_b\left(x\right)=\log_b\left(y\right) \iff x=y

This gives us two possible solutions x=-1 and x=4.

To test for extraneous solutions, we want to check if any solution falls outside the domain of the logarithms. The logarithmic terms in the original equation are \log_n\left(x+4\right), \log_n\left(x-2\right), and \log_n\left(x\right).

Any number between -4 and 2 will make the second term undefined, so the domain for this equation is \left(2, \infty\right).

We now need to check if either solution is extraneous. As -1 <2, it falls outside the domain of the function and is therefore extraneous.

x=4 is a valid solution.

We can check our solution by substituting x=4 into the equation:

Sketch the graph of y=7\log_{10}\left(x+3\right).

We can graph this equation by first completing a table of values and drawing the curve that passes through these points, or we can consider how the function f\left(x\right)=\log_{10}\left(x\right) has been transformed.

Considering each component separately, we can see that the graph of y=\log_{10}\left(x\right) is stretched vertically by a factor of 7 and translated 3 units to the left.

Remember that we apply stretches, compressions, and reflections before applying translations.

Graph y=x+2 on the same coordinate plane.

The graph of y=x+2 is a straight line with a slope of 1 and a y-intercept of 2.

Estimate the solutions to the equation, rounding to the nearest integer.

The solutions to the equation will be the points where the two graphs intersect.

Rounding to the nearest integer, we get the solutions x=-2 and x=4.

We can confirm that x=-2 is exactly a solution algebraically, as the left hand side of the equation becomes 7\log_{10}\left(1\right)= 0, and the right hand side becomes -2+2=0, as expected. We cannot do the same for x=4, which tells us that this solution must have been rounded.