solve log a (8x+5) = log a (4x+29)?

Answer 1

LOG a (8x+5) = LOG a (4x+29)

LOG (with a base of "a") = LOG (with a base of "a")

Therefore: # = #

8x +5 = 4x+29

8x - 4x + 5 = 4x - 4x + 29

4x + 5 = 0 + 29

4x + 5 = 29

4x + 5 - 5 = 29 - 5

4x + 0 = 24

4x = 24

4x / 4 = 24 / 4

x = 6

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1st RESTRICTION on "x" :

4x + 29 > 0
4x + 29 - 29 > 0 - 29
4x + 0 > -29
4x > -29
4x/4 > -29/4
x > -7.25

AND

2nd RESTRICTION on "x" :

8x + 5 > 0
8x + 5 - 5 > 0 - 5
8x + 0 > -5
8x > -5
8x/8 > -5/8
x > -5/8

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Take the larger of the two values for the "correct" RESTRICTION on "x".
This will make both statements TRUE for "x".

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i.e. FIRST we try a value where x > -7.25
Let x = -7

x > -7.25
-7 > -7.25 TRUE

x > -5/8
-7 > -5/8 FALSE

Since both statements for "x" are NOT TRUE,
we try the the 2nd RESTRICTION for "x".

We try a value x > -5/8
Let x = 0

x > -5/8
0 > -5/8 TRUE

x > -7.25
0 > -7.25 TRUE

Since both statements are TRUE for "x",
the final answer for "x" must be where: x > -5/8

THE SOLUTION:
x = 6

QUESTION:
Is 6 > -5/8 ?

ANSWER:
Yes.

CONCLUSION:
Therefore, x = 6 is a valid solution and is the solution to the question.

EXCEPTION:
If, however, either of the statements were FALSE,
and "x" was NOT greater than -5/8
Then, the answer to the question would be NO SOLUTION.

Answer 2

log a (8x+5)= log a (4x+29)
condition 1: a must differ from 1 and be more than 0
condition 2: 8x+5>0 <=> x>-5/8
solve:
log a (8x+5)=log a (4x+29)
<=> 8x+5=4x+29 and x>-5/8
<=> 8x-4x=29-5 and x>-5/8
<=> 4x=24 and x>-5/8
<=> x=6 and x>-5/8 (x=6 > -5/8, x is correct)
conclusion: x=6
good luck!

Answer 3

log a (8x+5) = log a (4x+29)
x must be greater than -5/8
<=> 8x+5 = 4x+29 and 01
<=> x =6 and 01

Answer 4

undertaking a million. because of the fact the bases are the two equivalent to a, in basic terms set 8x + 5 = 4x + 29. You get 4x = 24, so x = 6 undertaking 2. To isolate t, divide the two sides by making use of Pe^r. t = A / (Pe^r). undertaking 3. Substituting the final variety for I, D = 10 log( (5.4E-10)/1E-12)) I have been given 27.32

Answer 5

either u do this question directly by cancelling logs .such as:
log (8x+5)=log(4x+29)
=> 8x+5=4x+29
=> 8x+5=4x+29
=> 8x-4x=29-5
=> 4x=24
=>x=24/4
so, x=6

or using exponential form i.e.
8x+5 4x+29
e = e

=> (8x+5)-(4x+29)
e = 1

=> 4x-24 0
e = e
so, 4x-24=0 or x=6

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Answer 6

log a (8x+5) = log a (4x+29)
8x+5=4x+29
4x=24
x=6

Answer 7

log a (8x+5) = log a (4x+29)
8x+5=4x+29
8x-4x=29-5
4x=24
so x=6

Answer 8

8x+5=4x+29
4x=24
x=6

Answer 9

All those are correct!