How do you solve #logx-log2=1#?

Answer 1

x = 20

Using the following #color(blue)"laws of logarithms"#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(logx-logy=log(x/y))color(white)(a/a)|)))........ (A)# This applies to logarithms to any base.

#color(red)(|bar(ul(color(white)(a/a)color(black)(log_b a=nhArra=b^n)color(white)(a/a)|)))........ (B)#

#"Using (A) " logx-log2=log(x/2)#

A logarithm expressed as log x , usually indicates that the base is 10.

#"Using (B) " log_(10)(x/2)=1rArrx/2=10^1=10#

Thus #x/2=10rArrx=20#

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Answer 2

Lily Anderson

To solve the equation ( \log(x) - \log(2) = 1 ), you can use the properties of logarithms. Start by combining the logarithms using the quotient rule:

[ \log(x) - \log(2) = \log\left(\frac{x}{2}\right) ]

Now, rewrite the equation with the combined logarithm:

[ \log\left(\frac{x}{2}\right) = 1 ]

To eliminate the logarithm, exponentiate both sides of the equation with base 10:

[ 10^{\log\left(\frac{x}{2}\right)} = 10^1 ]

[ \frac{x}{2} = 10 ]

Now, solve for ( x ):

[ x = 2 \cdot 10 ]

[ x = 20 ]

So, the solution to the equation ( \log(x) - \log(2) = 1 ) is ( x = 20 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.