QuestionSolve for $x$ in these equations: a) $\log _{2} x+\log _{4} x=4.5$ b) $\log (x+2)+\log (x-1)=1$

Studdy Solution

STEP 1

Assumptions1. The base of the logarithm is greater than0 and not equal to1. The argument of the logarithm is greater than03. The logarithm properties are valid, such as $\log_a mn = \log_a m + \log_a n$ and $\log_a m^n = n \log_a m$
4. The equation is solvable for real values of $x$

STEP 2

First, we will solve the equation $\log{2} x+\log{4} x=4 \frac{1}{2}$ . We can simplify the equation by changing the base of the second logarithm to2. We know that $\log{b} a=\frac{\log{c} a}{\log{c} b}$ , so we can write $\log{4} x$ as $\frac{\log{2} x}{\log{2}4}$ .
$\log{2} x+\frac{\log{2} x}{\log{2}4}=4 \frac{1}{2}$

STEP 3

implify the equation by calculating $\log{2}$ . We know that $2^2=$ , so $\log{2}=2$ .
$\log{2} x+\frac{\log{2} x}{2}= \frac{1}{2}$

STEP 4

Multiply the whole equation by2 to get rid of the fraction.
$2\log{2} x+\log{2} x=9$

STEP 5

Combine the logarithms on the left side of the equation.
$3\log{2} x=9$

STEP 6

Divide both sides of the equation by3.
$\log{2} x=3$

STEP 7

To find the value of $x$ , we can rewrite the equation in exponential form. The base is2, the exponent is3, and the result is $x$ .
$x=2^3$

STEP 8

Calculate the value of $x$ .
$x=2^3=8$ So, the solution for the equation $\log{2} x+\log{4} x=4 \frac{1}{2}$ is $x=8$ .

STEP 9

Next, we will solve the equation $\log (x+2)+\log (x-)=$ . We can simplify the equation by using the logarithm property $\log_a mn = \log_a m + \log_a n$ .
$\log [(x+2)(x-)]=$

STEP 10

Expand the expression inside the logarithm.
$\log [x^2+x-2]=$

STEP 11

To find the value of $x$ , we can rewrite the equation in exponential form. The base is10 (since it's not specified), the exponent is, and the result is $x^+x-$ .
x^+x-=10^

STEP 12

implify the equation.
$x^2+x-2=10$

STEP 13

Subtract10 from both sides of the equation to set it equal to zero.
$x^2+x-12=0$

STEP 14

Factor the quadratic equation.
$(x-3)(x+4)=0$

STEP 15

Set each factor equal to zero and solve for $x$ .
$x-3=0 \quad or \quad x+4=0$

STEP 16

olve for $x$ .
$x=3 \quad or \quad x=-4$ However, we must check these solutions in the original equation, because the argument of a logarithm must be greater than zero.

STEP 17

Check $x=3$ in the original equation $\log (x+2)+\log (x-)=$ .
$\log (3+2)+\log (3-)=$

STEP 18

implify the equation.
$\log5+\log2=$

STEP 19

Combine the logarithms using the property $\log_a mn = \log_a m + \log_a n$ .
$\log10=1$ This is true, so $x=3$ is a valid solution.

STEP 20

Check $x=-4$ in the original equation $\log (x+)+\log (x-)=$ .
$\log (-4+)+\log (-4-)=$

STEP 21

implify the equation.
$\log (-)+\log (-5)=1$ This is not valid, because the argument of a logarithm must be greater than zero. So, $x=-4$ is not a solution.
So, the solution for the equation $\log (x+)+\log (x-1)=1$ is $x=3$ .

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