Question about Texas Instruments TI-83 Plus Calculator

# Log^-1(11/20) please solve step by step. - Texas Instruments TI-83 Plus Calculator

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Assuming you want the common antilog of 11/20:
2nd 10^x 1 1 / 2 0 ENTER

10^x is on the LOG key, to the left of the 7 key.

Posted on Nov 15, 2010

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Posted on Jan 02, 2017

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4726 seems to already be on base 10
it's 4*10^3 + 7*10^2 + 2*10^1 + 6*10^0

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### 2log(base4)x + log(base4)3=log(base4)x-log(base4)2

Hi,
It is easier to use the rules for logarithms to clean this equation before trying to solve it. And the calculator is no help here.
Let me represent the logs as follows log b in base a = log(a,b). The first argument is the base.

2*log(4,x) - log(4,3)= log(4,x)- log(4,2)

Here you have two similar terms 2*log(4,x) on the left and log(4,x) on the right. Gather them both on the same side, say the left. You have

2*log(4,x) -log(4,x) +log(4,3) = log(4,x)-log(4,x) -log(4,2)

which is equivalent to log(4,x) +log(4,3) = -log(4,2)

Gather the constant terms on the right, and use log(a) +log(b)=log(a*b)

log(4,x) = -log(4,2) -log(4,3) = - { log(4,2)+log(4,3)} = -log(4,2*3)

You get log(4,x)= -log(4,6).

Take the constant term to the left, while changing its sign in the process. On the right remains 0, which is also the log of 1 in any base So log(4,x)+log(4,6) =0 =log(4,1)

Use the rule log(a) +log(b) = log(a*b) valid in any base, to obtain
log(4,6x)=log(4,1) which is equivalent to 6x=1. The solution is x=1/6

There are a few other shortcuts to get the solution but I preferred to take a pedestrian, step by step, approach.

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