Question

The value of x, if log, 1250 + log, 80 = X, bases are 10
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Answer 1

Answer:

The value of x in  $log_{10} 1250 + log_{10} 80$ = x is 5

Step-by-step explanation:

Given: log 1250 + log,80 = X, bases are 10

To find: The value of x

Solution:

log 1250 + log,80 = X, bases are 10

We have to determine x values

We know that log ab=log a +log b

$log_{10} 1250 + log_{10} 80 = log_{10} 125 * 10 + log_{10} 8 *10$

$log_{10} 125 * 10 + log_{10} 8 *10 = log_{10} 125 +log_{10}10 + log_{10} 8 +log_{10}10$         (1)

⇒ $log_{10} 125 = log_{10} 5^{3}$

⇒ $log_{10} 5^{3} =$$3log5$

             =3 × 0.7

             = 2.1

$log_{10} 10 = 1$

$log_{10} 8 = log_{10} 2^{3}$

$log_{10} 2^{3}$= 3log 2

= 3 × 0.3

= 0.9

Sub the above values in eq  1

$log_{10} 125 +log_{10} 10+ log_{10} 8 +log_{10}10$

= 2.1 + 1 + 0.9 + 1

= 5

Final answer:

The value of x in  $log_{10} 1250 + log_{10} 80$ = x is 5

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Answer 2

Answer:

The required value of x is 5.

Step-by-step explanation:

Some rules of logarithm:-

• log a + log b = log($ab$)
• log a - log b = log($\frac{a}{b}$)
• $log_a(x)=b$ ⇒ $a^b=x$
• $e^{logx}=x$
• log($10^n$) = $n$
• log($a^b$) = b log a
• log10 = 1

Step 1 of 1

Given:

log1250 + log80 = x when base is 10, i.e.,

$log_{10}(1250)+log_{10}(80)=x$

To find:

The value of x.

Consider the given logarithm function as follows:

$log_{10}(1250)+log_{10}(80)=x$

Using the rules of logarithm, we have

⇒ $log_{10}(1250\times 80)=x$

⇒ $log_{10}(100000)=x$

⇒ $log_{10}(10^5)=x$

⇒ 5$log_{10}(10)=x$

As we know, the value of $log_{10}(10)$ is 1.

⇒ 5 × 1 = $x$

⇒ $x$ = 5

Final answer: The value of x is 5.

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