We are given an exponential equation. When bases are not the same, we can solve such equation by taking the logarithm of each side of the equation.
m=n ⇔ log m = log n
Note that in order to take their logarithms, both m and n must be positive numbers. In our case, we should first isolate the exponential term.
2*3^x=40.8 ⇔ 3^x=20.4
Let's now solve the equation using the Properties of Logarithms.
b This time we are given a polynomial equation. Let's begin by isolating the x-term. To do it we will divide the given equation by 3.
3x^4=27 ⇔ x^4=9Now, since both sides of the equation are positive we can take the fourth root of both sides to eliminate the power. Keep in mind that the root is even, so we have to consider positive and negative solutions.
We found two solutions of the given equation, x ≈ 1.732 and x ≈ - 1.732.
c To solve the given logarithmic equation we will rewrite it in exponential form using the definition of a logarithm.
log_b x=y ⇔ x= b^yThis definition tells us how to rewrite the logarithm equivalent of y in exponential form. The argument x is equal to b raised to the power of y.
log_5( 2x+1)=3 ⇔ 2x+1= 5^3
We can see above that 3 is the exponent to which 5 must be raised to obtain 2x+1. Now, let's solve our equation.
d Once again, we are given a logarithmic equation. However, this time we want to solve an equation involving more than one logarithm.
log(x)+log(2x)=5
Since the two logarithms are being added, we can use the Product Property of Logarithms.
log_b mn = log_b m + log_b n
Let's apply the above property.
Now, as in Part C we will rewrite the obtained equation in exponential form using the definition of a logarithm. Recall that if the base is not stated it is 10.
log_(10)( 2x^2)=5 ⇔ 2x^2 = 10^5
Let's solve it. Keep in mind that since x was an argument of a logarithm, we only need to consider positive solutions.
