Solve Logarithmic Equation: 2 Log to Base x of 5 + Log to Base 5 of x = Log 1000

[ll-Angel-ll]

Hi

I really need ur help

I was sick in the days that the teacher gave us the lesson

and I really really need ur help



this is the equation :


2 log{to base x} of 5 + log{to base 5} of x = log 1000


i know that log 1000 = 3


and 2 log{to base x} of 5 = log{to base x} of 25


so

log{to base x} of 25 + log{to base 5} of x = 3

and now I don't know how to continue

cause I don't know how to add different bases



please someone helps me

how can I continue and how can I add different bases ?!

HELP ME ..


​

 

[Dick]

Don't add different bases. Convert the base. Do you know:

log{base a}b=log{base c}b/log{base c}a.

Where c is anything. Use this to convert log{base x}25 to an expression involving log{base 5} only. I.e. put c=5 in the above.

 

[VietDao29]

Hi

I really need ur help

I was sick in the days that the teacher gave us the lesson

and I really really need ur help



this is the equation :


2 log{to base x} of 5 + log{to base 5} of x = log 1000


i know that log 1000 = 3​

Yup, so far, so good. :)

...
cause I don't know how to add different bases



please someone helps me

how can I continue and how can I add different bases ?!

HELP ME ..


​

Since you cannot add different bases. You need to convert it to only one base. The Change of Base Formula would did it:

\log_{a}b = \frac{\log_c b}{\log_c a}

This is what Dick told you.

From the above Formula, we can derive another 2 pretty nice, and also important Formulae:
1. \log_a b \times \log_b c = \log_a c

2. \log_a b = \frac{\log_b b}{\log_b a} = \frac{1}{\log_b a}

Remember the three formulae above, you may need it in Logarithmic Problems. :)

 

[ll-Angel-ll]

ok i think i got it

but don't laugh if it's wrong

2 log {base x} of 5 + log {base 5} of x = log 1000

1/( 2 log {base 5} of x ) + ( log {base 5} of x )/1 = 3

1/( 2 log {base 5} of x ) + (( log {base 5} of x )( 2 log {base 5} of x ))/( 2 log {base 5} of x ) = 3

so

( log {base 5} of x )/( 2 log {base 5} of x ) + 1 = 3

( log {base 5} of x )/( 2 log {base 5} of x ) = 3 -1

( log {base 5} of x )/( 2 log {base 5} of x ) =2

1/x = 2

so

x=1/2


am i right ?!

:shy:

​

 

[Dick]

Nope. Not quite. Let's call L=log{base 5}x, then yes, log{base x}5=1/L. But compare your first and second equations. How did the 2 move from the numerator to the denominator?

 

[ll-Angel-ll]

ok so u'r saying that what i did was correct

that 1/x = 2

but x doesn't equal 1/2

right ?!

​

 

[Dick]

No, I'm saying there are problems near the start. Your second equation is 1/(2L)+L=3 (again, let L=log{base 5}x). I'm trying to convince you that it should be 2/L+L=3.

 

[ll-Angel-ll]

ok so

2 log {base x} of 5 + log {base 5} of x = log 1000

2 / 2 log {base 5} of x + log {base 5} of x = 3

2 / log {base 5} of x^2 = 3

2 / 2 log {base 5} of x = 3

1 / log {base 5} of x = 3

log {base x} of 5 = 3

right ?!




​

 

[Dick]

Still wrong. You are doing some really funny looking algebra. How does

2 log{base x}5 become 2/(2 log{base 5}x)? It should be just

2/(log{base 5}x). Where is the extra 2 coming from?

 

[ll-Angel-ll]

ok i think i got it

2 log {base x} of 5 + log {base 5} of x = log 1000

[2/ log {base 5} of x] + log {base 5} of x = 3

ok then i should

ummm ..

[2 (log {base 5} of x) (log {base 5} of x)] / log {base 5} of x =3

2 (log {base 5} of x) =3

so

(log {base 5} of x^2) = 3

am i right this time ?!



​

 

[Dick]

Not right yet, but making progress. You now have 2/L+L=3. (Where L is your log{base 5}x. Can you solve 2/L+L=3 for L? I think the wordiness of your notation may be confusing you. Solve 2/L+L=3. It's the same equation. What's the first step?

 

[ll-Angel-ll]

cool I'm making progress

ok

2/L+L=3

first we multiply both sides by 1/2

so it will be

L+L = 3/2

2L = 3/2

so

L = 3/4


am i right ?!

:shy:

​

 

[Dick]

Sorry, but I think you need some serious practice on basic algebra.

(1/2)*(2/L+L)=(1/2)*(2/L)+(1/2)*L=1/L+L/2. That's the correct result - but it didn't simplify things much. What you really want to do is multiply both sides by L (to get rid of the L in the denominator). You are going to get a quadratic equation. Can you handle those?

 

[ll-Angel-ll]

i think the teacher was telling me the truth when she said I'm stupid

2 log {base x} of 5 + log {base 5} of x = log 1000

[2/ log {base 5} of x] + log {base 5} of x = 3

L = log {base 5} of x

2/L + L = 3

2+L^2 = 3L

L^2 - 3L + 2 = 0

( L-2 ) ( L -1 ) = 0

L = 2 or L = 1

----------------

what do i do now ?!


​

 

[Dick]

Now you are ready to collect your prize. L=log{base 5}x=1 or 2. Can you solve for x in each of those two cases?

 

[ll-Angel-ll]

ok

if

log {base 5} of x =1

then x = 5

and if

log {base 5} of x = 2

then x = 25

am i right ?!

:shy:

​

 

[Dick]

You win! Yes. Can you check that they both work in the original equation? I was serious about doing some algebra practice if you want to tackle these questions on your own.

 

[ll-Angel-ll]

thank u

thank u

thank u

i can't believe that i solved it

actually i couldn't solve it without u

thank u

thank u

thank u


ok and now i'll check

2 log {base x} of 5 + log {base 5} of x = log 1000

2 log {base 5} of 5 + log {base 5} of 5 = 3

2 * 1 + 1 = 3

2 + 1 = 3



========

2 log {base x} of 5 + log {base 5} of x = log 1000

2 log {base 25} of 5 + log {base 5} of 25 = 3

2 * 0.5 + 2 = 3

1 + 2 = 3



========

i'm so happy

and again

thank u

thank u

thank u

​