How do I solve and check this logarithm? (By hand and on calc.)?

Question

Hw do i solve the logarythm 2log8(2x-5)=4

(the 8 is the log and 2x-5 is not an exponent)

please tell me how to solve and check this by hand and calc.

thnx :)

Answer 1

First, I try to get rid of the coefficient 2 in front of the log. Divide by 2:

log8 (2x-5) = 2

Now, rewrite the log in "exponential form". That means the base of the log, 8, moves across the "=" and becomes the base of an exponential expression instead. The letters "log" simply disappear. We now have:
2x-5 = 8^2
2x-5=64
2x=69
x=34.5

To check, I would replace x in the original equation with 34.5.
2log8 (2x-5) = 2log8 [2(34.5)-5]
= 2log8 [69-5]
= 2log8 (64)

Here's the part where most calculators can't hang. It is extremely rare to find a calculator that can compute a base 8 log. Two ways through:

2log8 (64) = 2log8 (8*8)
=2[log8(8) + log8(8)] (using a log rule)
=2[1+1]
= 2*2
= 4

The other road, using your calculator, requires that you know a more complicated log rule--the "change of base" formula:

logb (a) = logc (a) / logc (b)

Where c is any valid base. If you choose c=10 (easiest), you get logb (a) = log(a)/log(b) because the log button on your calculator means log10.

Where were we? Oh yes: 2log8 (64). This can be expressed as 2log(64)/log(8) = 2*1.806179974/0.903089987 = 4.

Answer 2

2log 8 (2x-5)=4
Given the equation " 2 times the logarithm of (2x-5) to base 8 equals to 4 ", we need to solve for values of x that satisfy the equation.
Before we attempt to solve for x, do recall that for ' logarithm of b to base a ' [we use the notation " log a (b) " here] to be defined, b>0 and a>0, and a is not 1.

log 8 (2x-5) = 4 / 2
log 8 (2x-5) = 2
(8)^2 = 2x-5 {since y = a^x <==> x = log a (y) , where a is any positive number except 1}

2x-5 = 64
2x = 64 + 5
2x = 69
x = 69 / 2

To check our answer, put x = 69 / 2 into the left hand side of the equation :
2 log 8 { [ (2) ( 69 / 2 ) ] - 5 }
= 2 log 8 (64)
= 2 log 8 (8 ^ 2)
{ recall " log a (y) " is the power that a must have, in order to give y. So if y=a^2, then [ log a (y) ] = 2. Also, for any number k, k = k log a (a) = log a (a^k) }
= (2) (2)
= 4
Do note that since (2x-5) = 64, 64 > 0, 8 > 0, and the base of 8 is not equals to 1, therefore the left hand side of the question,
2log 8 (2x-5) , is defined.


To use a calculator, we must make use of the Change of Base Law.
Recall:
If a, b and c are positive numbers, and a is not 1, c is not 1, then
log a (b) = [ log c (b) ] / [ log c (a) ]

Since the question " 2log 8 (2x-5)=4" has base 8, which cannot be found in a general scientific calculator, we can change the base of 8 to either base 10 (using the button with the word "lg" or "log". Either of them is used to represent base 10 in most calculators.) or to base e (using the button with the word "ln", otherwise known as natural logarithm. The natural logarithm is the logarithm of something to the base e, where e is a certain mathematical constant approximately equal to 2.718281828459. ).

To change the base of 8 to base 10,
2log 8 (2x-5)
= 2 { [ lg (2x - 5) ] / [ lg (8) ] }

Using base 10, we can input the following into a calculator:
2 { [ lg (2 (69 / 2) - 5) ] / [ lg (8) ] }
and we will obtain 4.
Do be careful about the number and position of the brackets when attempting to input the expression into a calculator.


To change the base of 8 to base e,
2log 8 (2x-5)
= 2 { [ ln (2x - 5) ] / [ ln (8) ] }

Using base e, we can input the following into a calculator:
2 { [ ln (2 (69 / 2) - 5) ] / [ ln (8) ] }
and we will obtain 4.

Since both base 10 and base e produce an answer of 4, we have somewhat verified the Change of Base Law, albeit this is not rigorous enough.
In cases where the calculator's output is an error message, it could be due to either
1. wrong number of brackets
2. wrong positioning of brackets
3. range of values of a and b that does not satisfy the conditions " b>0 and a>0, and a is not 1 ", such that ' logarithm of b to base a ' [we use the notation " log a (b) " here] is not defined.
4. input of an expression different from the intended

Answer 3

The equation you're trying to solve is

2 times log, base 8, of (2x-5) = 4 The 8 is the base for your logarithm; it is not your logarithm.

Solution:

log, base 8, of (2x-5) = 2 from dividing each side by 2

convert log expression to an equivalent exponent expression

(In general log, base b, of (argument) = x
converts to b^x = argument. This is a fundamental idea for logarithms.)

8^2 = (2x-5) Solve: x = 69/2

Answer 4

Remember a log is saying " how many times must I multiply this base by itself to arrive at this number?" In your case your base is 8 and the number is (2x-5) and it was raised to the 4th power. So this means, that (2x-5) is equal to the base of 8 to the forth power.
You should be able to divide (2x-5) by the base of 8 four times.

Answer 5

2log8(2x - 5) = 4
log8(2x - 5) = 2
2x - 5 = 8^2 = 64
2x = 69
x = 34.5

2Log8(2*34.5 - 5) =
2Log8(69 - 5) =
2Log8(64) =
2*2 = 4

On a calculator:
[ 4 ] [ / ] [ 2 ] [ = ] [ MS ] [ 8 ] [ x^y ] [ MR ] [ = ] [ + ] [ 5 ]
[ = ] [ / 2 ] [ = ]

Answer 6

for calc graph both equations and find where they intersect
Y=2log8(2X-5)
Y=4
don't remember by hand lol

Answer 7

That really helped thanks!!