Evaluate and solve: log 5 + log 16 − 3log 2, 2log 3 − 2log 2, log10(2x−5)=1, log(x+2)+log(x−2)=log 60

Question

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let's evaluate these logarithmic expressions and solve for $x$ .

Evaluate:

$\log 5 + \log 16 - 3\log 2$

Apply the power rule of logarithms, $n \log a = \log a^n$ .
$\log 5 + \log 16 - \log 2^3$ $\log 5 + \log 16 - \log 8$
Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ .
$\log (5 \times 16) - \log 8$ $\log 80 - \log 8$
Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .
$\log \left(\frac{80}{8}\right)$ $\log 10$
Since the base of $\log$ is usually 10 when not specified, $\log_{10} 10 = 1$ .
$1$
The value is $\boxed{1}$ .
1. $2\log 3 - 2\log 2$
Apply the power rule of logarithms, $n \log a = \log a^n$ .
$\log 3^2 - \log 2^2$ $\log 9 - \log 4$
Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .
$\log \left(\frac{9}{4}\right)$
The value is $\boxed{\log \left(\frac{9}{4}\right)}$ .

Find the value of $x$ :
1. $\log_{10}(2x - 5) = 1$
Convert the logarithmic equation to an exponential equation.
If $\log_b a = c$ , then $b^c = a$ .
$10^1 = 2x - 5$ $10 = 2x - 5$
Solve the linear equation for $x$ .
$10 + 5 = 2x$ $15 = 2x$ $x = \frac{15}{2}$
The value of $x$ is $\boxed{\frac{15}{2}}$ .
1. $\log (x+2) + \log(x-2) = \log 60$
Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ , to the left side.
$\log ((x+2)(x-2)) = \log 60$
If $\log A = \log B$ , then $A=B$ .
Equate the arguments of the logarithms.
$(x+2)(x-2) = 60$
Expand the left side using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ , and solve the quadratic equation.
$x^2 - 4 = 60$ $x^2 = 60 + 4$ $x^2 = 64$ $x = \pm \sqrt{64}$ $x = \pm 8$
Check for valid solutions.
The argument of a logarithm must be positive.
For $x=8$ : $(8+2) = 10 > 0$ and $(8-2) = 6 > 0$ . This is a valid solution.
For $x=-8$ : $(-8+2) = -6 < 0$ . This is not a valid solution.
The value of $x$ is $\boxed{8}$ .
1. $\frac{\log 32}{\log 128} = x$
Express 32 and 128 as powers of the same base, which is 2.
$32 = 2^5$ $128 = 2^7$
Substitute these into the equation.
$x = \frac{\log 2^5}{\log 2^7}$
Apply the power rule of logarithms, $\log a^n = n \log a$ .
$x = \frac{5 \log 2}{7 \log 2}$
Cancel out $\log 2$ from the numerator and denominator.
$x = \frac{5}{7}$
The value of $x$ is $\boxed{\frac{5}{7}}$ .

Evaluate and solve: log 5 + log 16 - 3log 2, 2log 3 - 2log 2, log10(2x−5)=1, log(x+2)+log(x−2)=log 60

Step-by-step solution

Apply the power rule of logarithms, $n \log a = \log a^n$ .

Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ .

Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .

Since the base of $\log$ is usually 10 when not specified, $\log_{10} 10 = 1$ .

Apply the power rule of logarithms, $n \log a = \log a^n$ .

Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .

Convert the logarithmic equation to an exponential equation.

Solve the linear equation for $x$ .

Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ , to the left side.

If $\log A = \log B$ , then $A=B$ .

Expand the left side using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ , and solve the quadratic equation.

Check for valid solutions.

Express 32 and 128 as powers of the same base, which is 2.

Apply the power rule of logarithms, $\log a^n = n \log a$ .

Cancel out $\log 2$ from the numerator and denominator.

Still stuck on this one?

More Mathematics Questions

Evaluate and solve: log 5 + log 16 - 3log 2, 2log 3 - 2log 2, log10(2x−5)=1, log(x+2)+log(x−2)=log 60

Step-by-step solution

Apply the power rule of logarithms, nlog⁡a=log⁡ann \log a = \log a^nnloga=logan.

Apply the product rule of logarithms, log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log (ab)loga+logb=log(ab).

Apply the quotient rule of logarithms, log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right)loga−logb=log(ba​).

Since the base of log⁡\loglog is usually 10 when not specified, log⁡1010=1\log_{10} 10 = 1log10​10=1.

Apply the power rule of logarithms, nlog⁡a=log⁡ann \log a = \log a^nnloga=logan.

Apply the quotient rule of logarithms, log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right)loga−logb=log(ba​).

Convert the logarithmic equation to an exponential equation.

Solve the linear equation for xxx.

Apply the product rule of logarithms, log⁡a+log⁡b=log⁡(ab)\log a + \log b = \log (ab)loga+logb=log(ab), to the left side.

If log⁡A=log⁡B\log A = \log BlogA=logB, then A=BA=BA=B.

Expand the left side using the difference of squares formula (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2(a+b)(a−b)=a2−b2, and solve the quadratic equation.

Check for valid solutions.

Express 32 and 128 as powers of the same base, which is 2.

Apply the power rule of logarithms, log⁡an=nlog⁡a\log a^n = n \log alogan=nloga.

Cancel out log⁡2\log 2log2 from the numerator and denominator.

Still stuck on this one?

More Mathematics Questions

Apply the power rule of logarithms, $n \log a = \log a^n$ .

Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ .

Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .

Since the base of $\log$ is usually 10 when not specified, $\log_{10} 10 = 1$ .

Apply the power rule of logarithms, $n \log a = \log a^n$ .

Apply the quotient rule of logarithms, $\log a - \log b = \log \left(\frac{a}{b}\right)$ .

Solve the linear equation for $x$ .

Apply the product rule of logarithms, $\log a + \log b = \log (ab)$ , to the left side.

If $\log A = \log B$ , then $A=B$ .

Expand the left side using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ , and solve the quadratic equation.

Apply the power rule of logarithms, $\log a^n = n \log a$ .

Cancel out $\log 2$ from the numerator and denominator.