SOLUTION: Determine the value of ab if https://imgur.com/a/1u4d55h

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Question 1153085: Determine the value of ab if https://imgur.com/a/1u4d55h
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

log(8,a)+log(4,b^2)=5
log(8,b)+log(4,a^2)=7

log%288%2Ca%29%2Blog%284%2Cb%5E2%29=5.....eq.1
log%288%2Cb%29%2Blog%284%2Ca%5E2%29=7.....eq.2
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log%288%2Ca%29%2Blog%284%2Cb%5E2%29=5.....eq.1, solve for log%28a%29
log%288%2Ca%29=5-log%284%2Cb%5E2%29......change to base 10
log%28a%29%2Flog%288%29=5-log%28b%5E2%29%2Flog%284%29
log%28a%29=log%288%29%285-log%28b%5E2%29%2Flog%284%29%29
log%28a%29=3log%282%29%285%2A2log%282%29-2log%28b%29%29%2F2log%282%29....simplify,
log%28a%29=3cross%28log%282%29%29%285%2A2log%282%29-2log%28b%29%29%2F2cross%28log%282%29%29
log%28a%29=3%285%2A2log%282%29-2log%28b%29%29%2F2...simplify
log%28a%29=3%285%2Across%282%29log%282%29-cross%282%29log%28b%29%29%2Fcross%282%29
log%28a%29=3%285%2Alog%282%29-log%28b%29%29
log%28a%29=15log%282%29+-+3log%28b%29............eq.1a

go to
log%288%2Cb%29%2Blog%284%2Ca%5E2%29=7.....eq.2, do same as above

log%284%2Ca%5E2%29=7-log%288%2Cb%29
log%28a%5E2%29%2Flog%284%29=7-log%28b%29%2Flog%288%29
2log%28a%29=log%284%29%287-log%28b%29%2Flog%288%29%29
2log%28a%29=14log%282%29+-+%282log%28b%29%29%2F3
log%28a%29=7log%282%29-log%28b%29%2F3.........eq.2a

from eq.1a and eq.2a we have
15log%282%29+-+3log%28b%29=7log%282%29-log%28b%29%2F3
8log%282%29=9log%28b%29%2F3+-log%28b%29%2F3
8log%282%29=8log%28b%29%2F3....simplify
log%282%29=log%28b%29%2F3
3log%282%29=log%28b%29
log%282%5E3%29=log%28b%29
log%288%29=log%28b%29
=> b+=+8
go to
log%28a%29=15log%282%29+-+3log%28b%29............eq.1a, substitute b
log%28a%29=15log%282%29+-+3log%288%29
log%28a%29=15log%282%29+-+3log%282%5E3%29
log%28a%29=15log%282%29+-+9log%282%29
log%28a%29=6log%282%29
log%28a%29=log%282%5E6%29
a=2%5E6
a+=+64

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


log%288%2Ca%29%2Blog%284%2Cb%5E2%29+=+5
log%288%2Cb%29%2Blog%284%2Ca%5E2%29+=+7

Change the bases of each logarithm, knowing that 2%5E2=4 and 2%5E3=8.

%281%2F3%29log%282%2Ca%29%2B%281%2F2%29log%282%2Cb%5E2%29+=+5
%281%2F3%29log%282%2Cb%29%2B%281%2F2%29log%282%2Ca%5E2%29+=+7

Simplify using the exponent rule of logarithms.

%281%2F3%29log%282%2Ca%29%2Blog%282%2Cb%29+=+5
%281%2F3%29log%282%2Cb%29%2Blog%282%2Ca%29+=+7

Change the two equations to exponential form. (See below for an alternative method for solving the problem from this point.)

a%5E%281%2F3%29%2Ab+=+2%5E5
b%5E%281%2F3%29%2Aa+=+2%5E7

Multiply the two equations together:

%28ab%29%5E%284%2F3%29+=+2%5E12
ab+=+2%5E9+=+512

ANSWER: ab = 512

It's more work; but you could also solve the problem by using elimination to solve the pair of equations in the two unknowns, log%282%2Ca%29 and log%282%2Cb%29.

3log%282%2Cb%29%2Blog%282%2Ca%29+=+15
%281%2F3%29log%282%2Cb%29%2Blog%282%2Ca%29+=+7
%288%2F3%29log%282%2Cb%29+=+8
log%282%2Cb%29+=+3
b+=+2%5E3+=+8

3log%282%2Ca%29%2Blog%282%2Cb%29+=+21
%281%2F3%29log%282%2Ca%29%2Blog%282%2Cb%29+=+5
%288%2F3%29log%282%2Ca%29+=+16
log%282%2Ca%29+=+6
a+=+2%5E6+=+64